Differentiate the exponential function $f(x)= \frac{x^2e^x}{x^2+e^x}$

Question

$$f(x)= \frac{x^2e^x}{x^2+e^x}$$

Using product rule and quotient rule I computed

$$f'(x)=\frac{(x^2+e^x)e^x(x^2 + 2 x ) - x^2e^x(2x+e^x)}{(x^2+e^x)^2}$$

Is my computation correct so far?

Answer 1

Given $\dfrac{x^2e^x}{x^2+e^x}$

Apply the quotient rule: $\left(\dfrac uv\right)=\dfrac{u^{\prime\cdot}\cdot v-v^{\prime}\cdot u}{v^2}$ $$\dfrac{d}{dx}\left(\dfrac{x^2e^x}{x^2+e^x}\right)=\dfrac{\dfrac{d}{dx}(x^2e^x)(x^2+e^x)-\dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=\dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$

$$=\dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$

Edit:

$\dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule

By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$

Answer 2

If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have $$\ln f(x) = 2 \ln x + x - \ln (x^2 + e^x).$$ Differentiating with respect to $x$ gives $$\frac{f'(x)}{f(x)} = \frac{2}{x} + 1 - \frac{2x + e^x}{x^2 + e^x},$$ which, after some algebra and simplification, reduces to $$f'(x) = \frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$