Question

compute the derivative of the given function in two different ways.

g(x)=-2x^{3}(7x^{5})

a) use the product rule, (f(x)g(x) = f(x)cdot g(x)+f(x)cdot g(x)). (fill in each blank, then simplify.)

g(x)=(square)cdot(square)+(square)cdot(square)=square

b) use algebra first to simplify (g), then differentiate without the product rule.

g(x)=square

Explanation:

Step1: Identify f(x) and g(x) for product - rule

Let $f(x)=-2x^{3}$ and $g(x)=7x^{5}$.

Step2: Find f'(x) and g'(x)

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $f'(x)=-2\times3x^{2}=-6x^{2}$ and $g'(x)=7\times5x^{4}=35x^{4}$.

Step3: Apply the product rule

$g'(x)=f(x)\cdot g'(x)+f'(x)\cdot g(x)=(-2x^{3})\cdot(35x^{4})+(-6x^{2})\cdot(7x^{5})$.
$g'(x)=-70x^{7}-42x^{7}=-112x^{7}$.

Step4: Simplify g(x) first

$g(x)=-2x^{3}(7x^{5})=-14x^{8}$.

Step5: Differentiate the simplified g(x)

Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, $g'(x)=-14\times8x^{7}=-112x^{7}$.

Answer:

a) $g'(x)=(-2x^{3})\cdot(35x^{4})+(-6x^{2})\cdot(7x^{5})=-112x^{7}$
b) $g'(x)=-112x^{7}$