Answer:
f'(x) = 7x^4 (5 ln(x) + 1) + x^2
Explanation:
[tex]\sf Let \ f(x)=7x^5 ln(x) + \dfrac{1}{3} x^3[/tex]
Properties used while differentiating:
[tex]i) \quad \sf \dfrac{d}{dx} (ln(x)) = \dfrac{1}{x}[/tex]
[tex]ii) \quad \sf \dfrac{d}{dx} (x^n) = n(x)^{n-1}[/tex]
[tex]iii) \quad \sf \dfrac{d}{dx}(uv)=u \dfrac{dv}{dx}+v \dfrac{du}{dx}[/tex]
Begin solving:
f'(x) =
[tex]\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\:+\:\dfrac{1}{3}\:\:x^3\right)[/tex]
[tex]\Longrightarrow \sf \dfrac{d}{dx}\:\left(7x^5\:ln\left(x\right)\right) + \dfrac{d}{dx}\left(\dfrac{1}{3}x^3\right)[/tex]
[tex]\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)[/tex]
[tex]\Longrightarrow \sf 5(7x^{5-1})\:\times\ ln(x) + \dfrac{1}{x} \:\times\ 7x^5 \ + x^2[/tex]
[tex]\Longrightarrow \sf 35x^{4} ln(x) + \ 7x^4 \ + x^2[/tex]
[tex]\Longrightarrow \sf 7x^4(5 ln(x) + 1) \ + x^2[/tex]
