Question

Finding second Derivatives In Exercise,find the second derivate.
f(x) = (1 + 2x)e4x

Answer 1

`(d^2)/(dx^2)\left(\left(2x+1\right)e^(4x)\right)=32e^(4x)x+32e^(4x)`.

Explanation:

To find the second derivative of the function
`f(x)=\left(2 x + 1\right) e^(4 x)\right)` you must:

Step 1. Find the first derivative
`(d)/(dx)\left(\left(1+2x\right)e^(4x)\right)`

`\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\\\\f=1+2x,\:g=e^(4x)`

`(d)/(dx)\left(\left(1+2x\right)e^(4x)\right)=(d)/(dx)\left(1+2x\right)e^(4x)+(d)/(dx)\left(e^(4x)\right)\left(1+2x\right)`

`(d)/(dx)\left(1+2x\right)=2\\\\(d)/(dx)\left(e^(4x)\right)=e^(4x)\cdot \:4`

`(d)/(dx)\left(\left(1+2x\right)e^(4x)\right)=2e^(4x)+e^(4x)\cdot \:4\left(1+2x\right)\\\\(d)/(dx)\left(\left(1+2x\right)e^(4x)\right)=6e^(4x)+8e^(4x)x`

Step 2. Find the second derivative
`(d)/(dx) (6e^(4x)+8e^(4x)x)`

`\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\(d)/(dx)\left(6e^(4x)+8e^(4x)x\right)=(d)/(dx)\left(6e^(4x)\right)+(d)/(dx)\left(8e^(4x)x\right)\\\\=24e^(4x)+8\left(4e^(4x)x+e^(4x)\right)\\\\32e^(4x)x+32e^(4x)`

`(d^2)/(dx^2)\left(\left(2x+1\right)e^(4x)\right)=32e^(4x)x+32e^(4x)`.