Respuesta :
Answer:
[tex]\frac{6x^2+3}{2x^3+3x}[/tex]
Step-by-step explanation:
You need to apply the chain rule here.
There are few other requirements:
You will need to know how to differentiate [tex]\ln(u)[/tex].
You will need to know how to differentiate polynomials as well.
So here are some rules we will be applying:
Assume [tex]u=u(x) \text{ and } v=v(x)[/tex]
[tex]\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}[/tex]
[tex]\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}[/tex]
[tex]\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}[/tex]
[tex]\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}[/tex]
Those appear to be really all we need.
Let's do it:
[tex]\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)[/tex]
[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))[/tex]
[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})[/tex]
[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))[/tex]
[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)[/tex]
[tex]\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}[/tex]
I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.
Your answer is [tex]\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}[/tex].
