Help
Options
Didn't know it?
click below
click below
Knew it?
click below
click below
Don't Know
Remaining cards (0)
Know
retry
shuffle
restart
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
Normal Size Small Size show me how
Maths Y13 Pure
Maths Autumn Y13
| Question | Answer |
|---|---|
| What is the range of a function? | All the values that can come out of a function (acceptable outputs) |
| What is the domain of a function? | All the values that go in (acceptable inputs) [domaIN] |
| What does the fancy Z with superscript + mean? | Positive integers not including zero (aka counting numbers) |
| What does the fancy R mean | Real numbers |
| What does the fancy Q mean | Rational numbers |
| How to add conditions to set notation | Commas (however its said by Ms Wade that you cant do epsilon R when > or < involved as it cant be any real number - feels off though) |
| What is one-to-one mapping | Every value on the domain is mapped onto a unique value in the range (e.g. general rule 2x + 5) - no crossover of graph (e.g. just straight line) |
| What is one-to-many mapping | A single value in the domain is mapped onto 2+ *unique* values in the range (e.g. two lines on graph) |
| What is many-to-one mapping | 2+ values in the domain are mapped onto the same value in the range (e.g. x^3 graph) |
| What is many-to-many mapping | Each value in the domain mapped onto non-unique multiple values (e.g. circle graph) |
| Useful tool for seeing stretches | Writing out how (x,y) would be affected |
| Drawing weird stretches on sin/cos/tan graphs | Dont stress about the scaling - the labelling is the important part |
| Multiple transformations upon the same dimension | The y transformations affect the whole equation as normal The x transformations affect only the x - not everything in the brackets. |
| Remember with scaling x by 3 | That means it becomes 1/3 x |
| 'One way stretch' | Just means stretch |
| Do the compound angle formulae have to be in radians | No |
| How to prove tan(A+B) | Divide by cosA cosB Look where you're going with the 1 to work it out |
| When working with simplifying sin, cos, etc with real values | Can turn e.g. sin(pi/6) into nice numbers first |
| If Q asks for a tanx value | Dont give the x value |
| cotangent (cot), cosecant (cosec), secant (sec) way to remember | 3rd letter is the function its the inverse of Being the inverse of a function means that its the opposite flip for sohcahtoa, and can also just be the value of tanx then ^-1 |
| Alternate method to triangle for cosx = 3/5 being converted to sinx | Sub in to sin^2 + cos^2 = 1 |
| When finding secondary solutions to a function | CHECK WITH GRAPH ALWAYS FOR WORKING MARKS YOU NEED TO DRAW IT ALSO NEED TO CHECK IF IT EVEN HAS SECONDARY SOLUTIONS - DOES IT HAVE THE SAME Y VALUE FOR MULTIPLE X POINTS IN A PERIOD |
| Generally when simplifying both sides by dividing by smth when doing sin/cos/tan | Reconsider dividing by sin as it'll just lead to messy cots rather than tans - still works but kinda annoying |
| Dividing by a variable | NEVER divide by a variable (unless forming an identity?) - loses solutions. Factorise out instead |
| Sin curve translation | A sin curve translated parallel to the x axis and stretched parallel to the y axis will have the form rsin(x + alpha) for some alpha and r. Using compound angles, can be written as asinx + bcosx |
| e.g. Put 3sinx + 4coxx in the form Rsin(x+a) giving a in degrees to 1dp Finding R | Expand Rsin(x+a) = Rsinxcosa + Rcosxsina Compare coefficients: 3 = Rcosa 4 = Rsina Square both coefficient comparisons, then add. Factorise out into R^2(cos^2 + sin^2) = R^2 = 9 + 16 R = 5 |
| e.g. Put 3sinx + 4coxx in the form Rsin(x+a) giving a in degrees to 1dp Finding a and giving equation | Using the fact Rsina/Rcosa = tana 4/3 = tana a = 53.1 degrees (principal easiest if they dont mind which solution) Sub into equation |
| Anything with e.g. theta - 68 degrees as the solution to a cos^-1 | CHANGE THE RANGE Work out the solutions before adjusting back to pure theta, changing all the solutions |
| When you're doing the R method of changing double angle formulas and they dont give you an end point | Just choose one - they all work as versions of each other |
| cos^-1(x) | Doesnt mean cosx to the power of -1 because history, instead meaning arccos Reciprocal doesnt mean inverse for trig |
| For when you have a split range where no number is in both | Use 'or' or union. Comma and 'and' not valid |
| For easy tan drawing | Draw asymptotes first |
| Multiple equals sign in one line | Can apparently only do one per line or incorrect |
| Finding the new identities | Divide the sin^2 + cos^2 identity by sin^2 or cos^2 Always derive to make sure its correct |
| e.g. 1/tantheta = 0 | Cant divide by zero so no reciprocal solution. Either show graph and select points when cot(theta) = 0 [which is where the asymptotes of tan are] or do cos/sin = 0, so cos = 0 |
| Double angle proof step 1 | Suppose we had a line length 1 projected at angle A+B above horizontal. Angle A forms right angled triangle, then the angle B forms a right angle triangle with the hypotenuse equalling 1. Line connecting from base to very top point is line XY = sin(A+B) |
| Double angle proof step 2 | Add a new right angled triangle to the gap between the previous two. XY = the two vertical lines to the right of it added together |
| Double angle proof step 3 | Find lengths of top triangle with hypotenuse 1. O = sinB. A = cosB |
| Double angle proof step 4 | Find lengths of two smaller vertical lines, each in terms of both sin and cos. Add together and equal it to sin(A+B) |
| Using double angle proof for sin(A-B) | sin(A-B) = sin(A+(-B)) = sinAcos(-B) + cosAsin(-B) Draw graphs to show cos(-B) = cosB and sin(-B) = -sinB Sub in |
| Using double angle proof for cos(A+B) | cos(A+B) just shifted by pi/2 of sin(A+B) cos(A+B) = sin(pi/2 - (A+B)) = sin((pi/2 - A) + (-B)) = sin(pi/2 - A) cos(-B) + cos(pi/2 - A) sin (-B) = cosA cosB - sinA sinB |
| Inverse trig identities | Not the same as ^-1. We have to restrict domain so one-to-one (only type of function we can inverse). Reflected in y = x. Careful that they dont end at the intersection: continue on to pi/2 after meeting at 1,1 |
| Given y = arccosx, express arcsinx in terms of y | y = cos^-1(x) x = cosy x = sin(pi/2 - y) [this is just true: reflection and shift of 90 degrees ] pi/2 - y = arcsinx |
| sin(5(x-30)) if translation happens first | Translate by 150/0 then stretch horizontally by 1/5 |
| Map graph of y = 2(x+1)^2 + 5 onto y = x^2 | ONTO x^2 so opposite direction |
| How to do inverse function | f(x) = 2x + 1 Let 2x + 1 = y [this is because reflected in y=x] y = 2x + 1 (y-1)/2 = x. then turn the y into an x and the x into an f(x) [domain and range being swapped] |
| Domain and range of inverse function | Swapped |
| Self-inverse functions | If and only if ff(x) = x |
| What does f:x ---> mean | f(x) = |
| What functions can have an inverse | A function can have an inverse if and only if it is a 1-to-1 function |
| Modulus functions | Sketches rlly useful Can do with just common sense of flipping gradient, but better to get it into the form I2x+1I < 7 (modulus only thing on that side), then turn that into -7 < 2x+1 < 7. Check graph to see if split < > |
| 'Solve the equation' | x values |
| If(x)I | Flipped in x axis |
| f(IxI) | As if the x value inputs were the positive: reflect the positive x value graph in the y axis |
| Section of curve with d^2y/dx^2 > 0 | Concave upwards (i.e. if points at 2 ends connected by line, curve below that line Vice versa for <0 |
| Points of inflection | When d^2y/dx^2 = 0, where it switches from concave upwards to concave downwards. Points of inflection can be stationary but not always. Not all d^2y/dx^2 = 0 are points of inflection |
| Which graphs have points of inflection | Every cubic has a point of inflection. |
| d^2y/dx^2 | Show change in concavity - useful when they want to know if inflection or not by going x = 5.1 and x = 4.9 to show the change about 5 If switches from +ve to -ve, is a point of inflection rather than max/min |
| Conditions for max point | dy/dx = 0 d^2y/dx^2 < 0 (or can be zero sometimes) |
| Conditions for min point | dy/dx = 0 d^2y/dx^2 > 0 (or can be zero sometimes) |
| Conditions for point of inflection | Can be 0 or not for dy/dx d^2y/dx^2 = 0 (when you get this =0, check both dy/dx both sides to see if actually inflection or not) |
| When d^2y/dx^2 switches from +ve to -ve | Not max/min - is a point of inflection. Can have stationary or non-stationary points of inflection (stationary means gradient = 0) |
| Chain rule | dy/dx = dy/du x du/dx Subbing in u for the contents of brackets e.g. (2x + 1)^10 du/dx means u differentiated with respect to x (the u isolated on one side) Remember to sub back in the u as the original always |
| If asked to find the other point of inflection | Still ALWAYS prove its a point of inflection |
| Rate always means | something/dt |
| e.g. radius r cm of circular ripple increasing at a rate of 8cms^1. At what rate is the area A increasing when radius is 25cm? | dA/dt (what we're looking for) = dr/dt (given in Q) x dA/dr (what the format must have via chain rule logic) A = pi*r^2 dA/dr = 2*r*pi Solve |
| How to get db/dx from a known value of dx/db | ^-1 |
| When you get an answer with multiple variables | Check in Q for a relationship between the two to get in terms of one variable. |
| e.g. sphere volume decreasing at rate of 60. Find rate of decrease of SA when volume is 800 | dV/dt = 60 V = 4/3*pi*r^2 dV/dr = 4*pi*r^2 A = 4*pi*r^2 dA/dr = 8*pi*r Looking for dA/dt dA/dt = dV/dt * dr/dV * dA/dr (look for nice diagonals cancelling) 800 = 4/3*pi*r^3 r = 5.8 Solve |
| Differentiated form of y = e^kx | dy/dx = ke^kx If the x is a function, do with chain rule |
| Differentiated form of lnx | 1/x Can only differentiate if a pure x is there - if it's a function, do with chain rule |
| Differentiate y = lnx^2 | =2lnx = 2/x Can do your normal log rearrangings first before differentiating |
| Product rule | If y = uv, then dy/dx = u*dv/dx + v*du/dx (differentiate each and multiply by the other one, then add) |
| Quotient rule | For when y = u/v In formula booklet |
| Removable solutions with logs | Negative logs - watch for them |
| Showing function is increasing | Must have line "f'x > 0" at some point |
| Vector notation | Underline them |
| Discriminant < 0 | So no *****REAL***** roots |
| Getting quick magnitude vector form | OPTN -> ANGLE -> POL(-3, 7) or OPTN -> ANGLE -> REC(sqrt58, 113 degrees) |
| Giving velocity answer | Give in VECTOR form. Works to do 4i + 10j - k |
| Easiest way to prove right angled | Pythagoras |
| Proving the 4th point (H) location on a parallelogram | Need to include: EF (with arrow above) = vector form EF = HG = vector GH = vector H = answer |
| If modulus after the flip is gradient = 3 | Gradient before the flip was -3 |
| Check for if =< is valid too | |
| Specify 'minimum point' for reasoning | |
| e.g. sec(2t) | MUST chain rule it |
| dy/dx(sinx) = | cosx |
| dy/dx(cosx) = | -sinx |
| dy/dx(tanx) = | sec^2(x) [can technically work out with quotient] |
| dy/dx(1) | 0 I'm not sure why this is here either honestly |
| Useful equation for union | P(AuB) = P(A) + P(B) - P(AnB) If mutually exclusive, = P(A) + P(B) because P(AnB) = 0 |
| Equation for given | Probability of A given B = P(A l B) = P(AnB)/(P(B) Limiting set to just P(B) then seeing what proportion of that is also P(A) If independent, P(A l B) = P(A) [just work it out] Super careful to do right way around if (B l A). Given under |
| Exhaustive | P(quehoawub) = 1 |
| What to always do for probability question | Draw venn diagram - really nice for working out the different sector values, and is especially vital for non independent |
| For the denominator in the given equation | It uses the chance of it that variable in any branch |
| Sample spaces | Should draw for 'X of 2 numbers' questions For given: number of applicable values in the restricted space/restricted sample space |
| Does AnB' = (AnB)' | No |
| 'In context' | Say what sets A and B stand for |
| (AuBuC)' = | A'nB'nC' (also flips sign when expanding) |
| A few pointers for hypothesis testing | Don't assume significance is 5% - check Do both tails in calculator to choose most significant (most thorough is to check significance of > and < in the bcd menu when doing calc). Don't have to show why you chose - just write the smallest tail. |
| Rules if A and B are independent | P(AnB) = P(A) x P(B) P(A l B) = P(A) |
| Rules if A and B are mutually exclusive | P(AnB) = 0 P(AuB) = P(A) + P(B) |
| If you see 'given that' | IMMEDIATELY write out law for conditional probability (like even before finishing reading Q) Vital for preventing mistakes and working marks |
| If you see the words 'are independent' | IMMEDIATELY write out laws for independence (like even before finishing reading Q) Vital for preventing mistakes and working marks |
| nCr = | (n over r) = n!/r!(n-r)! |
| (-1)! | -1 * -2 * -3 * -4 ....... * -infinity |
| Doing choose for negative and fraction numbers | (-2)!/(3!)(-2-3)! Write out -2 * -3 * -4 etc for the negative ! on the top and bottom, going a few past where they cancel out and ending both on the same number for clarity. Then do ellipses. Then can cancel equal sequences |
| e.g. binomial expansion of 1/(1+x) | = (1+x)^-1 Write out for each expansion: -1C0 * 1 * x^0 = 1 * 1 * 1 = 1 -1C1 * 1 * x^1 -1C2 * 1 * x^2 etc Work out the chooses with the formula from before (ALWAYS WRITE IT) |
| Binomial expansion of e.g. (3 + 15x)^-4 | Can't do the weird binomial expansion unless one of the terms is 1. So do: [3(1 + 5x)]^-4 = 3^-4 * (1+5x)^-4 Also remember not just x for the expansion |
| 'Valid' expansions | i.e. converging function For e.g. (1 + 5x), the magnitude of 5x must be less than one for the value to converge to a value. Happens as x gets progressively smaller. Just set the magnitude of the not-1 part as less than 1, then solve for x |
| e.g. sqrt(1+x)/sqrt(1-x) | = (1+x)^0.5 * (1-x)^-0.5 Then binomially expand each separately to the amount of xs the Q asks = (binomial expansion 1) * (binomial expansion 2) = long list going up to x^4 roughly = short list only going up to x^2 (whatever asked in Q) |
| 'write the equations' of possible lines | Not x = 0, pi, 2*pi Must be: x = 0 x = pi x = 2pi |
| sec(2theta) + tan(2theta) | Remember can express tan as sin/cos to give common denominator |
| Hypothesis testing don't forget | X squiggle B |
| Partial fractions e.g. [5x]/[(x+4)(x-1)] | 5x/(x+4)(x-1) = A/(x+4) + B/(x-1) 5x = A(x-1) + B(x+4) Can sub in any value, so at x = -4, A(-5) = -20 etc |
| Partial fractions with more than 2 denoms e.g. [x+8]/[(x-2)(2x+1)(x+3)] | = A/(x-2) + B/(2x+1) + C/(x+3) x + 8 = A(2x+1)(x+3) + B...... etc Multiply each numerator by BOTH other denoms |
| Partial fractions with indices on the denoms e.g. (1+2x)/(x-1)^2 | Dont try to just split into equal denoms because then you can just add them. Instead so they have different denoms initially but can still be turned into the correct denom, do = A/(x-1) + B/(x-1)^2 A(x-1) + B = 1 + 2x |
| Partial fractions with indices on the denoms and multiple denoms e.g. (9x-14)/[(x+4)(x-1)^2] | = A/(x+4) + B/(x-1) + C/(x-1)^2 9x - 14 = A(x-1)^2 + B(x+4)(x-1) + C(x+4) to give everything the right denom they need weird multiplying |
| e.g. Express (2x^4 - 5x^3 - 11x^2 + 7x - 26)/[(x+2)(x-4)] in the form Ax^2 + Bx + C + D/(x+2) + E/(x-4) | The A,B,C show us we can divide Divide using quotient 2x^2 - x + 3. GIves remainder 5x -2 2x^2 - x + 3 + (5x-2)/(x^2 - 2x - 8) Partial fraction the end part. |
| Remainders (Y12 flashcard) | Remember: If the matrix method gives -15 when it should be -3, then the remainder is +12 (what you need to make it normal). |
| e.g. Find first 3 terms of binomial expansion of (7x-1)/(3-x)(1+3x) | Partial fractions give 2/(3-x) - 1/(1+3x) 2(3-x)^-1 = 2/3(1-(1/3)x)^-1 Then do the two binomial expansions and just subtract them |
| If random variable X is normally distributed, we write | X squiggle N(μ, σ^2) μ = mean σ = standard deviation |
| Normal distribution averages | Symmetrical so mean = mode = median |
| +- 1 standard deviation | Contains 68% of the data (in formula booklet i think) |
| +- 2 standard deviation | Contains 95% of the data |
| +- 3 standard deviation | Contains 99.7% of the data |
| +- 5 standard deviation | Contains all of the data (pretty much) |
| Ways to prove not normally distributed | Skews and discrete (positive skew means bump to the left) Discrete not normal but can approximate |
| What is the y axis on the normal distribution | Probability density (like freq density but the probability of it lying within the area. Total area under curve is therefore 1) |
| Straight line of the probability density | *Uniform* distribution |
| When using distribution menu on calc | ALWAYS view value with the OPTN F1 - it truncates otherwise |
| Using distribution menu for inverse normal | Can edit the p value to get the x value(s) |
| e.g. Find c such that P(26<y<c) = 0.2 | Draw graph. P(Y<c) - P(Y<26) = 0.36 Use calc for P(Y<26) |
| e.g. Find d such that P(25-d < y < 25+d) = 0.6 | Draw graph. The outer bounds are 0.2 each. then P(Y<25+d) = 0.8 easy on calc |
| Probability | Not % |
| Standard normal distribution | If X squiggle N(μ, σ^2) and Z squiggle N(0,1) Then P(X < x) is the same as P(Z < (x - μ)/σ). The transformation of x into (x - μ)/σ is called standardising |
| P(Z < (x - μ)/σ) [all of it] sometimes written as | ϕ((x - μ)/σ) Phi is a functionish thing meaning P(Z <. Allows for more coherent working when inversing. ONLY FOR STANDARD |
| e.g. standardise X squiggle N(10, 2^2) P(X < 6) | X squiggle N(0,1) 6-10/2 = -2 P(Z<-2) = P(X<6) |
| When to do standardsising | Good for when mean and/or standard deviation are unknown Also good for comparisons - is just showing amount of standard deviations above/below the mean of zero. P(1 < Z < 2) means is between 1 and 2 standard deviations above |
| e.g. Length may be modelled by normal distribution with standard deviation = 3. 15% of components longer than 16cm. Find the mean. | X squiggle N(μ, 3^2) P(X > 16) = 0.15 P(X < 16) = 0.85 (because inverse does up to a point) Z squiggle N(0,1) P(Z < (16-μ)/3) = 0.85 16-μ/3 = ϕ^-1(0.85) = 1.036 [making sure to use (0,1)] etc |
| ϕ^-1(0.8) | Means area UP TO the result of this is 0.8. i.e. Only does P(X< ) Same style thing for normal inverse |
| What step to do after finding the mean with the standard normal | Do a plausibility check: for this question its about 1 standard deviation below 16cm, which makes sense because the ϕ^-1(0.85) is 1.036, which is close to one standard deviation above mean. |
| Normal distribution points of inflection | Points 1 standard deviation above and below mean |
| Normal approximations | Discrete values. Inclusive vs exclusive matters |
| Requirements for Normal approximations | Steps in possible values are small compared to standard deviation. Continuity corrections are made |
| Continuity corrects examples | (106 <= X <= 110) integers becomes (105.5 <= X <= 110.5) continuous 'less than 120' is P(X<119.5) [so it doesnt include the 120 itself] 'at least 110' is P(X >= 109.5) to catch the ones that round up to 110 so is 1 - P(X<109.5) |
| Binomial to normal approx | Bi(n,p) -> N(np, σ^2), where σ = sqrt(np(1-p)) |
| Distribution of sample means | For a random sample of size n taken from a random variable X, the sample mean X (with bar above) is approximately normally distributed with Xbar squiggle N(μ,(σ^2)/n) |
| Sample means | e.g. average of 3 numbers between 1 and 2 is a sample mean. Then tons of these sample means are made. We expect the sample means to vary about the population mean μ (mean doesnt change from the mean of the random variable) |
| Standard deviation of Xbar | sqrt(σ^2/n) = σ/sqrt(n) |
| Increasing sample size effect on sample means | Increasing sample size means increasing amount of random values drawn to make each sample. Sample means xbar (lower case for each one I think is the convention) will be closer to true mean μ, so standard deviation of Xbar reduces |
| Ways to prove normally distributed | Symmetry Almost all data within 3 standard deviations of mean (give values of upper and lower bounds) 95% of data within 2 standard deviations from mean (give upper and lower) Continuous |
| X and Xbar | Dont confuse them. X is each randomly chosen item. Xbar is the mean length of each sample. X might not be normally distributed but Xbar will be. |
| Normal hypothesis testing e.g. Rods believed to be 30cm, with standard deviation of 1. A sample of 20 rods gives mean of 29.5 Set-up of answer | Only talking about one sample. Hypotheses always in term of population parameter e.g. mean H0: μ = 30 H1: μ < 30 Let X be the length of the rod |
| Normal hypothesis testing e.g. Rods believed to be 30cm, with standard deviation of 1. A sample of 20 rods gives mean of 29.5 Answer | Assuming H0, X squiggle N(30,1^2) so Xbar squiggle N(30, 1^2/20) -> Xbar squiggle N(30, (sqrt0.05)^2) Use the μ and σ values for P(Xbar < 29.5) = 0.127 (table given in Q. The p value is the prob of the observed value or more extreme) Conclude |
| Why we did (sqrt0.05)^2 | To remind us of std dev not variance - ALWAYS do |
| Using critical region for normal hypothesis testing | Use inverse normal of the significance level to find the critical region then compare to the value. Critical regions good for if testing multiple samples. If looking for upper region of 10%, do inverse of 0.95 |
| How to state critical values | Critical region is Xbar < 232.7 or Xbar > 247.3 (have to be separated because no value does both) |
| Using critical region | Must prove with BOTH regions |
| 'Sample standard deviation' | Watch out for sample parameters rather than population |
| When can we use sample standard deviation as an estimate for the whole population | If the population standard deviation is not given and n is large enough (30+) Mention in your answer We still divide by sqrt(n) to find the standard deviation of the sample means |
| Function of binomial vs normal hypothesis test | Binomial is testing for a probability/proportion. Normal is testing for a population mean using a sample |
| r value | +ve r value means +ve gradient of lobf. aka 'product moment correlation coefficient' PMCC |
| r value hypothesis tests | H0 is always r = 0. They give you the critical values to do it with - essentially just choose the right tail |
| PMCC vs SRCC | PMCC is the r value, but the SRCC (R means Rank) is replacing the raw data with its rank in the data set |
| How to use SRCC | Replace raw data. Calculate PMCC using these ranks. Interpret the coefficient as rho = 0.86 for the original data and draw conclusions about association rho doesnt tell us about correlation - tells us if the data is strongly *associated* |
| How to adjust SRCC for repeat data points | Need to be adjusted to data isnt pushed out of sync |
| r = 1 | Perfectly straight line through origin (not necessarily y = x) |
| rho = 1 | Must be a *strictly* increasing function |
| r vs rho | r value for PMCC. r for the rank correlation (the correlation between the rank numbers, but doesnt tell us about the data set) rho for the SRCC of the original data |
| If you have the equation of the lobf | Dont estimate by drawing on graph, sub in instead |
| For something like an exponential graph, should r be used? | Probably not - would be too low. Ranks likely more reasonable. |
| If hypothesis says 'positive correlation' | Do SRCC and use rho for hypothesis ??? I dont know what this note means - check with someone smart if its correct |
| For chain rule of e.g. 1/5x | Dont just sub in u and then immediately unpack it treating it like an x. Do proper chain rule with dy/du |
| Implicit differentiation | Unlike explicit, doesnt have y as subject e.g. y^2 - 3xy = x^2 We must differentiate each term separately Remember constant terms *all* become zero |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x - splitting into terms | d(y^2)/dx - d(3xy)/dx = d(x^2)/dx |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x d(y^2)/dx = | d(y^2)/dy * dy/dx = 2y * dy/dx We cant differentiate y with respect to x - no sense. But we can do this by differentiating wrt y and then 'cancelling out' the dy |
| e.g. y^2 - 3xy = x^2 differentiated with respects to x d(3xy)/dx = | Product rule of 3x and y 3y + 3x * dy/dx [dy/dy just equals 1] |
| e.g. differentiate y = a^x | lny = lna^x lny = xlna 1/y * dy/dx = lna dy/dx = ylna y = a^x dy/dx = a^x*lna |
| Differentiation rule for a^kx | Original * loga * k |
Created by:
Pyrogearos2