Question

differentiate the given function.
y = x(x^5 + 2)^4
\frac{d}{dx}x(x^5 + 2)^4=\square

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = x$ and $v=(x^{5}+2)^{4}$. First, find $u^\prime$ and $v^\prime$. The derivative of $u = x$ with respect to $x$ is $u^\prime=\frac{d}{dx}(x)=1$.

Step2: Apply chain - rule to find $v^\prime$

Let $t=x^{5}+2$, so $v = t^{4}$. By the chain - rule, $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. The derivative of $v$ with respect to $t$ is $\frac{dv}{dt}=\frac{d}{dt}(t^{4}) = 4t^{3}$, and the derivative of $t$ with respect to $x$ is $\frac{dt}{dx}=\frac{d}{dx}(x^{5}+2)=5x^{4}$. Then $v^\prime = 4(x^{5}+2)^{3}\cdot5x^{4}=20x^{4}(x^{5}+2)^{3}$.

Step3: Calculate $y^\prime$

Using the product - rule $y^\prime=u^\prime v+uv^\prime$, substitute $u = x$, $u^\prime = 1$, $v=(x^{5}+2)^{4}$, and $v^\prime=20x^{4}(x^{5}+2)^{3}$ into it. We get $y^\prime=(x^{5}+2)^{4}+x\cdot20x^{4}(x^{5}+2)^{3}=(x^{5}+2)^{4}+20x^{5}(x^{5}+2)^{3}$.

Answer:

$(x^{5}+2)^{4}+20x^{5}(x^{5}+2)^{3}$