suppose y=(x^3 + 7x)(4x^2 + 9). here are two different ways to compute the derivative of this function.\n a…

Explanation:

Step1: Expand the product

[ \begin{align*} y&=(x^{3}+7x)(4x^{2}+9)\ &=x^{3}\times4x^{2}+x^{3}\times9 + 7x\times4x^{2}+7x\times9\ &=4x^{5}+9x^{3}+28x^{3}+63x\ &=4x^{5}+37x^{3}+63x \end{align*} ]

Step2: Differentiate the expanded function

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have: [ \begin{align*} y'&=\frac{d}{dx}(4x^{5}+37x^{3}+63x)\ &=4\times5x^{4}+37\times3x^{2}+63\times1\ &=20x^{4}+111x^{2}+63 \end{align*} ]

Step3: Apply the product rule

Let $f(x)=x^{3}+7x$ and $g(x)=4x^{2}+9$. Then $f'(x)=3x^{2}+7$ and $g'(x)=8x$. By the product rule $y'=f'g + fg'=(3x^{2}+7)(4x^{2}+9)+(x^{3}+7x)(8x)$.

Step4: Expand and simplify the product - rule result

[ \begin{align*} y'&=(3x^{2}+7)(4x^{2}+9)+(x^{3}+7x)(8x)\ &=3x^{2}\times4x^{2}+3x^{2}\times9+7\times4x^{2}+7\times9+x^{3}\times8x+7x\times8x\ &=12x^{4}+27x^{2}+28x^{2}+63 + 8x^{4}+56x^{2}\ &=(12x^{4}+8x^{4})+(27x^{2}+28x^{2}+56x^{2})+63\ &=20x^{4}+111x^{2}+63 \end{align*} ]

Answer:

a. $y = 4x^{5}+37x^{3}+63x$, $y'=20x^{4}+111x^{2}+63$ b. $y'=(3x^{2}+7)(4x^{2}+9)+(x^{3}+7x)(8x)$, $y'=20x^{4}+111x^{2}+63$