find the derivative of the following function.\nh(x) = (4x^7 + 4x)(4x^3 + 2x^2 + 2)

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=f(x)g(x)$, then $h^\prime(x)=f^\prime(x)g(x)+f(x)g^\prime(x)$. Let $f(x)=4x^{7}+4x$ and $g(x)=4x^{3}+2x^{2}+2$.

Step2: Find $f^\prime(x)$

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $f(x)=4x^{7}+4x$, we have $f^\prime(x)=\frac{d}{dx}(4x^{7})+\frac{d}{dx}(4x)=28x^{6}+4$.

Step3: Find $g^\prime(x)$

Using the power - rule, for $g(x)=4x^{3}+2x^{2}+2$, we have $g^\prime(x)=\frac{d}{dx}(4x^{3})+\frac{d}{dx}(2x^{2})+\frac{d}{dx}(2)=12x^{2}+4x+0 = 12x^{2}+4x$.

Step4: Calculate $h^\prime(x)$

$h^\prime(x)=(28x^{6}+4)(4x^{3}+2x^{2}+2)+(4x^{7}+4x)(12x^{2}+4x)$ Expand the first part: $(28x^{6}+4)(4x^{3}+2x^{2}+2)=28x^{6}(4x^{3}+2x^{2}+2)+4(4x^{3}+2x^{2}+2)=112x^{9}+56x^{8}+56x^{6}+16x^{3}+8x^{2}+8$. Expand the second part: $(4x^{7}+4x)(12x^{2}+4x)=4x^{7}(12x^{2}+4x)+4x(12x^{2}+4x)=48x^{9}+16x^{8}+48x^{3}+16x^{2}$. Combine like terms: $h^\prime(x)=(112x^{9}+48x^{9})+(56x^{8}+16x^{8})+56x^{6}+(16x^{3}+48x^{3})+(8x^{2}+16x^{2})+8$ $h^\prime(x)=160x^{9}+72x^{8}+56x^{6}+64x^{3}+24x^{2}+8$.

Answer:

$h^\prime(x)=160x^{9}+72x^{8}+56x^{6}+64x^{3}+24x^{2}+8$