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

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

Step-by-step explanation:

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

Properties used while differentiating:

`i) \quad \sf (d)/(dx) (ln(x)) = (1)/(x)`

`ii) \quad \sf (d)/(dx) (x^n) = n(x)^(n-1)`

`iii) \quad \sf (d)/(dx)(uv)=u (dv)/(dx)+v (du)/(dx)`

Begin solving:

f'(x) =

`\Longrightarrow \sf (d)/(dx)\:\left(7x^5\:ln\left(x\right)\:+\:(1)/(3)\:\:x^3\right)`

`\Longrightarrow \sf (d)/(dx)\:\left(7x^5\:ln\left(x\right)\right) + (d)/(dx)\left((1)/(3)x^3\right)`

`\Longrightarrow \sf (d)/(dx)(7x^5)\:\ * ln\left(x\right) + (d)/(dx)(ln(x)) * \ 7x^5 + \left(3\:* (1)/(3)x^(3-1)\right)`

`\Longrightarrow \sf 5(7x^(5-1))\:*\ ln(x) + (1)/(x) \:*\ 7x^5 \ + x^2`

`\Longrightarrow \sf 35x^(4) ln(x) + \ 7x^4 \ + x^2`

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