differentiate the following function. y = x(x³ + 3)⁴ d/dxx(x³ + 3)⁴ = □

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 $v=(x^{3}+3)^{4}$. 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 $t=x^{3}+3$, then $v = t^{4}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. We know that $\frac{dv}{dt}=\frac{d}{dt}(t^{4}) = 4t^{3}$ and $\frac{dt}{dx}=\frac{d}{dx}(x^{3}+3)=3x^{2}$. Substituting $t = x^{3}+3$ back in, we get $v^\prime=4(x^{3}+3)^{3}\cdot3x^{2}=12x^{2}(x^{3}+3)^{3}$.

Step3: Calculate $y^\prime$

Using the product - rule $y^\prime=u^\prime v+uv^\prime$, we substitute $u = x$, $u^\prime = 1$, $v=(x^{3}+3)^{4}$, and $v^\prime=12x^{2}(x^{3}+3)^{3}$: [ \begin{align*} y^\prime&=1\cdot(x^{3}+3)^{4}+x\cdot12x^{2}(x^{3}+3)^{3}\ &=(x^{3}+3)^{4}+12x^{3}(x^{3}+3)^{3}\ &=(x^{3}+3)^{3}[(x^{3}+3)+12x^{3}]\ &=(x^{3}+3)^{3}(13x^{3}+3) \end{align*} ]

Answer:

$(x^{3}+3)^{3}(13x^{3}+3)$