Differentiate using the chain rule:
d/du [ln(u)] d/dx[2x^3+3x]
derivative of ln(u) = 1/u
1/u d/dx[2x^3+3x]
1/2x^3+3x d/dx[2x^3+3x]
Differentiate
(6x^2+3) 1/2x^3+3x
Simplify
Dy/dx = 3(2x^2+1) / x(2x^2 +3)
Answer:
[tex]\frac{(6x^2+3)}{(2x^3+3x)}[/tex]
Step-by-step explanation:
[tex]\frac{dy}{dx}[/tex] [tex]ln(x)=[/tex] [tex]\frac{1}{x}[/tex] .
Therefore:
[tex]\frac{dy}{dx}[/tex] [tex]ln(2x^{3}+3x) = \frac{1}{2x^{3}+3x} *6x^2+3[/tex]
[tex]6x^2+3[/tex] is multiplied because of the chain rule.
