Question

Differentiate the following functions (a), (b), (c).?

Answer 1

`f'(x)=6x+12xln(4x)`
`g'(x)=(-2cos(2x)*e^(-2x)+2sin(2x)(e^(-2x)-1))/cos^2(2x)`
`h'(t)=(3t^2+2cos(2t))/(2sqrt(t^3+sin(2t)))`

Explanation:

Function A
We'll use the product rule. We can see:
`d/dx(6x^2)=6x*2=12x`
`d/dx(ln(4x))=1/(4x)*4=1/x` (by the chain rule)

Now we plug these into the product rule:
`d/dx(6x^2ln(4x))=(6x^2)/x+12xln(4x)=6x+12xln(4x)`

Function B
For this one we use the quotient rule. We can see:
`d/dx(cos(2x))=-2sin(2x)` (by the chain rule)
`d/dx(e^(-2x)-1)=-2e^(-2x)` (by the chain rule)

Now we plug this into the quotient rule:
`d/dx((e^(-2x)-1)/cos(2x))=(-2cos(2x)*e^(-2x)-(-2sin(2x)(e^(-2x)-1)))/cos^2(2x)`
`=(-2cos(2x)*e^(-2x)+2sin(2x)(e^(-2x)-1))/cos^2(2x)`

Function C
This is a chain rule application. To make it very clear, I will let `u=t^3+sin(2t)`

By the chain rule, we get:
`d/(du)(sqrt(u))*d/(dt)(t^3+sin(2t))=1/(2sqrtu)(3t^2+2*cos(2t))`

Resubstituting, we get:
`(3t^2+2cos(2t))/(2sqrt(t^3+sin(2t)))`