differentiate the following function\n\n$y = x(x^{2}+1)^{4}$\n\n$\frac{d}{dx}x(x^{2}+1)^{4}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = x$ and $v=(x^{2}+1)^{4}$. $\frac{d}{dx}[x(x^{2}+1)^{4}]=x\frac{d}{dx}[(x^{2}+1)^{4}]+(x^{2}+1)^{4}\frac{d}{dx}(x)$

Step2: Differentiate $u = x$

$\frac{d}{dx}(x)=1$

Step3: Differentiate $v=(x^{2}+1)^{4}$ using chain - rule

Let $t=x^{2}+1$, then $v = t^{4}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. $\frac{dv}{dt}=4t^{3}$ and $\frac{dt}{dx}=2x$. So $\frac{dv}{dx}=4(x^{2}+1)^{3}\cdot2x = 8x(x^{2}+1)^{3}$

Step4: Substitute back into product - rule formula

$x\cdot8x(x^{2}+1)^{3}+(x^{2}+1)^{4}\cdot1$ $=8x^{2}(x^{2}+1)^{3}+(x^{2}+1)^{4}$ $=(x^{2}+1)^{3}(8x^{2}+x^{2}+1)$ $=(x^{2}+1)^{3}(9x^{2}+1)$

Answer:

$(x^{2}+1)^{3}(9x^{2}+1)$