Differentiation C3

if y = cosx, then dy/dx = . . . ?

- sinx

differentiate cos½x

-½sin½x don't forget the coefficient of x!

How do you differentiate with expressions like 3xe^4x

1) Use the product rule 2) u = 3x v = e^4x u' = 3 v' = 4e^4x y' = uv' + vu' y' = (3x)(4e^4x) + (e^4x)(3) factor out e^4x y' = 3e^4x(4x + 1)

How do you find the gradient of a curve at a certain point? if f(x) = 4x² - 8x + 3 find the gradient of the curve at the point (1/2, 0 )

1) differentiate with respect to x f'(x) = 8x - 8 2) plug the x coordinate into f'(x) f'(1/2) = 8(1/2) - 8 gradient is -4

What is the quotient rule? How do you use it?

1) write down u with u' underneath it 2) write don v with v' underneath it 3) use the formula! 4) done

differentiate with respect to x: (2 + x)³(2 − x)³

1)First, ignore the powers and multiply the brackets together. (4 - x²)³ now use the chain rule. By inspection: (3)(-2x)( 4 - x²)³ f'(x) = -6x( 4 - x²)³

Find the range of values of x for which f(x) = 12lnx - x^(3/2) is an increasing funtion

It's an increasing function when the gradient is greater than 0 1) differentiate 2) set f'(x) > 0 3) solve for x

if y = sin3x, then dy/dx = . . .?

by the chain rule . . . . dy/dx = f'(x)cos(f(x)) dy/dx = 3cos3x

differentiate with respect to x (1 − 5x + x³)⁴

by the chain rule: u = 1 - 5x + x³ u' = -5 + 3x² y = u⁴ y' = 4u³ dy/dx = u'y' = (-5 + 3x²)(4u³) = 4( -5 + 3x² )( 1 - 5x + x³ )³

f(x) = x( 1 + x )lnx find f'(x)

consolidate into one bracket f(x) = (x + x²)lnx u = (x + x²) u' = 2x v = lnx v' = 1/x dy/dx = uv' + vu' = (x + x²)(1/x) + (1+ 2x)lnx = 1 + x + (1 + 2x)lnx

if y = sinx, then dy/dx = . . ?

cosx

For questions of type: T = 20 + 60e^−kt Find the rate of decrease of temperature when t = 40 ,

differentiate with respect to T dT/dt they often have tricky expressions in the power involving ln, etc. So calculate this part and use decimals eg. = 60 × (−0.07167)e^−0.07167t

Find the equation for the tangent to the curve C at the point with y coordinate 3 x = y³ - 4y²

differentiate with respect to y dx/dy = 3y² - 8y dx/dy when y = 3 .. . . 3(3)² - 8(3) = 3 gradient is dy / dx, NOT dx/dy, so gradient is 1/3 NOT 3 (do this bit like with C12) y - y₁ = m(x - x₁) y - 3 = (1/3)(x - (- 9)) y = x/3 + 6

what is f'(x) if f(x) is e^4

e^4 differential of e is e

for questions type: Curve C has equation f(x) = xsecx f'(x) = 1 + xtanx draw two suitable graphs to deduce the number of stationary points that f(x) has in the interval: 0 ≤ x ≤ 2π

f'(x) = 1 + xtanx stationary points where: 1 + xtanx = 0 xtanx = -1 tanx = -1/x not solvable with algebra: ∴ draw graph of: y = tanx and y = - 1/x number of points of intersection shows how many stationary points there are

differentiate e^2x

f'(x) = f'(x)(f(x) = 2(e^2x) =2e^2x

How can you do the chain rule by inspection?

if y = [f(x)]^n , dy/dx = nf'( x )[ f(x) ]^n-1 1) bring down the exponent n 2) multiply by the differential of the bracket nf'(x) 3) multiply by f(x) (without the exponent) nf'(x)[f(x)] 4)apply the exponent - 1 to the last bracket 5) nf'(x)[f(x)]^n-1

what is the formula of the gradient function?

if y = ax^n, then dy/dx = nax^n-1 or if f(x) = ax^n, then f'(x) = nax^n-1

if y = ln((x^2) +1), what is the differential?

if y = ln(fx), then dy/dx = f'(x)/f(x) 1) chain rule let u = (x^2) + 1 so that y = ln(u) u' = 2x y' = 1/u (ln rule) dy/dx = dy/du x du/dx = 1/u x 2x = 2x/u = 2x/((x^2) + 1) which equals f'(x) / f(x) (as the formula shows)

What is an easy way to remember the product rule?

if y = uv, then dy/dx = u(dv/dx) + v(du/dx) 1) write down u with du/dx underneath it 2) write down v with dv/dx underneath it (use chain rule if needed) 3) use the formula dy/dx = u(dv/dx) + v(du/dx) 4) simplify tip: you need to use the chain rule, doing it by inspection is much quicker

What is the purpose of differentiation?

it gives the gradient of the graph at point x

if y = sin²x, then dy/dx = . . . ?

sin²x = (sinx)² by the chain rule dy/dx = (2)(cosx)(sinx) = 2cosxsinx

What is the chain rule and how do you use it?

where y is a function of u and u is a function of x 1) let u = everything in the bracket 2) let y = (u)^a, where a is the exponent of everything in the bracket 3) write .the chain rule (as seen in the image) 4) work out dy/du 5) work out du/dx 6) calculate using the chain rule 7) Sub out u 8) done!

if y = lnx what is the differntial?

y' = 1/x