Question

Differentiate the function.
h(x) = e^x^4+ In(x)

Answer 1

`h'(x) = 4e^(4x) + (1)/(x)`

Explanation:

Given the function:

`h(x) = e^(x^4) + \ln(x)`

we can find its derivative using:

• the sum/difference rule ...
  `\left[\frac{}{}f(x) \pm g(x)\frac{}{}\right]' = f'(x) \pm g'(x)`
• the chain rule ...
  `\left[\frac{}{}f(x)^n\frac{}{}\right]' = n \left[\frac{}{}f(x)\frac{}{}\right]^(n\,-\,1) \cdot f'(x)`
• the common derivatives ...
  `\left[\,e^x\,\right]' = e^x` ...
  `\left[\frac{}{}\ln(x)\frac{}{}\right]' = (1)/(x)`

First, we can apply the sum/difference rule.

`h(x) = e^(x^4) + \ln(x)`

`h'(x) = \left[\frac{}{}e^(x^4)\frac{}{}\right]' + \left[\frac{}{}\ln(x)\frac{}{}\right]'`

Next, we can simplify the derivative of
`\ln(x)`.

`h'(x) = \left[\frac{}{}e^(x^4)\frac{}{}\right]' + (1)/(x)`

Then, we can simplify the derivative of the first term using the chain rule.

`h'(x) = \left[\frac{}{}4\,[\,e^x\,]^((4\, -\, 1))\cdot e^x\frac{}{}\right] + (1)/(x)`

Finally, this can be simplified using the properties of exponents.

`\boxed{h'(x) = 4e^(4x) + (1)/(x)}`