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

##### 1 Answer

#### Explanation:

*Finding the first derivative:*

Use the chain rule (many times).

**First Issue:** the sine function.

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

**Second Issue:** the square root.

#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 *negative sine* function and have the

#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)#