Find the derivative of $F(x) = x^7 \ln(x^3 e^{3x^2 -8})$

Question

Find $F'(x)$ for $$F(x) = x^7 \ln(x^3 e^{3x^2 -8})$$

Here is what I have so far:

$$(7x^6)(3x^2)$$

Answer 1

Hint:

$$\Big[f(x)g(h(x))\Big]' = f'(x)g(h(x)) + f(x)g'(h(x))h'(x)$$

Take $f(x) = x^7$, $g(h(x)) = \ln (x^3e^{3x^2 - 8})$.

Answer 2

Hint: Recall the product rule: $$\frac{d}{dx}\left(f(x)g(x)\right)=f'(x)g(x)+f(x)g'(x).$$ Identify $f(x)=x^7$ and $g(x)=\ln(x^3e^{3x^2-8})$. Then for $g'(x)$ you need to use the chain rule ($\dfrac{d}{dx}[g(p(x))]=g'(p(x))p'(x).$) and again product rule.

Answer 3

simplifying the given term $$\ln(x^3e^{3x^2-8})=\ln(x^3)+(3x^2-8)\ln(e)=3\ln(x)+3x^2-8$$ thus we get $$3x^7\ln(x)+3x^9-8x^7$$ and the first derivative is given by $$21x^6\ln(x)+3x^7 \cdot\frac{1}{x}+27x^8-56x^6$$