Question

Let f(x)=7x^5 ln(x) + 1/3 x^3


f’(x)=


I’m stuck on this calculus problem. If anyone could help me work this out. I would appreciate if you showed all your steps.

Answer 1

Answer:

f'(x) = 7x^4 (5 ln(x) + 1) + x^2

Explanation:

$\sf Let \ f(x)=7x^5 ln(x) + \dfrac{1}{3} x^3$

Properties used while differentiating:

$i) \quad \sf \dfrac{d}{dx} (ln(x)) = \dfrac{1}{x}$

$ii) \quad \sf \dfrac{d}{dx} (x^n) = n(x)^{n-1}$

$iii) \quad \sf \dfrac{d}{dx}(uv)=u \dfrac{dv}{dx}+v \dfrac{du}{dx}$

Begin solving:

f'(x) =

$\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\:+\:\dfrac{1}{3}\:\:x^3\right)$

$\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\right) + \dfrac{d}{dx}\left(\dfrac{1}{3}x^3\right)$

$\Longrightarrow \sf \dfrac{d}{dx}(7x^5)\:\ \times ln\left(x\right) + \dfrac{d}{dx}(ln(x)) \times \ 7x^5 + \left(3\:\times \dfrac{1}{3}x^{3-1}\right)$

$\Longrightarrow \sf 5(7x^{5-1})\:\times\ ln(x) + \dfrac{1}{x} \:\times\ 7x^5 \ + x^2$

$\Longrightarrow \sf 35x^{4} ln(x) + \ 7x^4 \ + x^2$

$\Longrightarrow \sf 7x^4(5 ln(x) + 1) \ + x^2$