find the derivative of the function.\n$f(x)=(x^{3}-10x)(4x^{2}+3x)$\n$f(x)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=x^{3}-10x$ and $v = 4x^{2}+3x$.

Step2: Find $u^\prime$

Differentiate $u=x^{3}-10x$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. So, $u^\prime=\frac{d}{dx}(x^{3}-10x)=3x^{2}-10$.

Step3: Find $v^\prime$

Differentiate $v = 4x^{2}+3x$ using the power - rule. So, $v^\prime=\frac{d}{dx}(4x^{2}+3x)=8x + 3$.

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

Using the product - rule $f^\prime(x)=u^\prime v+uv^\prime=(3x^{2}-10)(4x^{2}+3x)+(x^{3}-10x)(8x + 3)$. Expand the expressions: [ \begin{align*} &(3x^{2}-10)(4x^{2}+3x)+(x^{3}-10x)(8x + 3)\ =&3x^{2}(4x^{2}+3x)-10(4x^{2}+3x)+x^{3}(8x + 3)-10x(8x + 3)\ =&12x^{4}+9x^{3}-40x^{2}-30x + 8x^{4}+3x^{3}-80x^{2}-30x\ =&(12x^{4}+8x^{4})+(9x^{3}+3x^{3})+(-40x^{2}-80x^{2})+(-30x-30x)\ =&20x^{4}+12x^{3}-120x^{2}-60x \end{align*} ]

Answer:

$20x^{4}+12x^{3}-120x^{2}-60x$