How do you differentiate #f(x)=x(1+e^(x^2))^(1/5)# using the chain rule?

Question

William Abdalla · Accepted Answer

#f'(x)=(5+5e^(x^2)+2x^2e^(x^2))/(5(1+e^(x^2))^(4/5))#

First, use the product rule:

#f'(x)=(1+e^(x^2))^(1/5)color(blue)(d/dx[x])+xcolor(green)(d/dx[(1+e^(x^2))^(1/5)]#

Find each derivative separately.

#color(blue)(d/dx[x]=1#

The following requires heavy use of the chain rule:

#color(green)(d/dx[(1+e^(x^2))^(1/5)])=1/5(1+e^(x^2))^(-4/5)color(red)(d/dx[1+e^(x^2)]#

#color(red)(d/dx[1+e^(x^2)])=e^(x^2)d/dx[x^2]=color(red)(2xe^(x^2)#

Plug this back in.

#color(green)(d/dx[(1+e^(x^2))^(1/5)])=(2xe^(x^2)(1+e^(x^2))^(-4/5))/5=color(green)((2xe^(x^2))/(5(1+e^(x^2))^(4/5))#

Finally, plug these in to find #f'(x)#.

#f'(x)=(1+e^(x^2))^(1/5)+(2x^2e^(x^2))/(5(1+e^(x^2))^(4/5))#

Multiply the first term for a common denominator.

#f'(x)=(5+5e^(x^2)+2x^2e^(x^2))/(5(1+e^(x^2))^(4/5))#

Emma Aldridge · Answer

undefined To differentiate $ f(x) = x(1 + e^{x^2})^{1/5} $ using the chain rule, we proceed as follows:

1. Identify the outer function and the inner function.
   Outer function: $ u = u(v) = v^{1/5} $
   Inner function: $ v = v(x) = 1 + e^{x^2} $

2. Differentiate the outer function with respect to its variable.
   $ \frac{du}{dv} = \frac{1}{5}v^{-4/5} $

3. Differentiate the inner function with respect to its variable.
   $ \frac{dv}{dx} = \frac{d}{dx}(1 + e^{x^2}) = 0 + e^{x^2} \cdot \frac{d}{dx}(x^2) = 2xe^{x^2} $

4. Apply the chain rule: Multiply the derivatives of the outer and inner functions.
   $ \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = \frac{1}{5}(1 + e^{x^2})^{-4/5} \cdot 2xe^{x^2} $

5. Simplify the expression if necessary.