Question

What is the second derivative of `f(x)= sin(sqrt(3x-7))`?

Answer 1

`f''(x)=(-9(sqrt(3x-7)sin(sqrt(3x-7))+cos(sqrt(3x-7))))/(4(3x-7)^(3/2)`

Explanation:

Finding the first derivative:

Use the chain rule (many times).

First Issue: the sine function. `d/dx[sin(u)]=cos(u)*u'`.

`f'(x)=cos(sqrt(3x-7))*d/dx[sqrt(3x-7)]`

Second Issue: the square root. `d/dx[sqrtu]=d/dx[u^(1/2)]=1/2u^(-1/2)*u'=(u')/(2sqrtu)`

`f'(x)=cos(sqrt(3x-7))*3/(2sqrt(3x-7))=color(blue)((3cos(sqrt(3x-7)))/(2sqrt(3x-7))`

Finding the second derivative:

Use the quotient rule (and more chain rule).

`f''(x)=(2sqrt(3x-7)color(green)(d/dx[3cos(sqrt(3x-7))])-3cos(sqrt(3x-7))color(red)(d/dx[2sqrt(3x-7)]))/(2sqrt(3x-7))^2`

Find each internal derivative separately:

`d/dx[3cos(sqrt(3x-7))]`

This will be almost identical to finding `d/dx[sqrt(3x-7)]`, except that it will turn into a negative sine function and have the `3` multiplied by it. We could do the work, but why redo something we've already essentially done?

`color(green)(d/dx[3cos(sqrt(3x-7))]=(-9sin(sqrt(3x-7)))/(2sqrt(3x-7))`

The other derivative is

`d/dx[2sqrt(3x-7)]`

This, again, will be twice what we determined earlier, since we've basically already done this differentiation.

`color(red)(d/dx[2sqrt(3x-7)]=3/sqrt(3x-7)`

Plug these back in.

`f''(x)=((2sqrt(3x-7)(-9sin(sqrt(3x-7))))/(2sqrt(3x-7))-(9cos(sqrt(3x-7)))/sqrt(3x-7))/(4(3x-7))`

Further simplification yields:

`f''(x)=(-9(sqrt(3x-7)sin(sqrt(3x-7))+cos(sqrt(3x-7))))/(4(3x-7)^(3/2)`