Question

Find dy du , du dx , and dy dx .

Answer 1

Answer:

a) $\frac{dy}{du}=5(x^2 +1)^4$, $\frac{du}{dx}=2x$, $\frac{dy}{dx}=10x(x^2+1)^4$  

b) $\frac{dy}{du}=4(4x^2-x+6)^3$, $\frac{du}{dx}=8x-1$, $\frac{dy}{dx}=4(8x-1)(4x^2-2+6)^3$  

Step-by-step explanation:

We can use the chain rule in the following form: is u=u(x) is a differentiable function depending on x and y=y(u) is a differentiable function depending on u, then $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$.

a) $\frac{dy}{du}=\frac{d}{du} (u^5)=5u^4=5(x^2 +1)^4$ from the power rule.  

$\frac{du}{dx}=\frac{d}{dx} (x^2 +1)=2x$.

From the previous parts and the chain rule, $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=5(x^2 +1)^4(2x)=10x(x^2+1)^4$  

b) $\frac{dy}{du}=\frac{d}{du} (u^4)=4u^3=4(4x^2-x+6)^3$  

$\frac{du}{dx}=\frac{d}{dx} (4x^2-x+6)=8x-1$ from the power and sum rules.

Then, $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}=4(4x^2 -x+6)^3(8x-1)=4(8x-1)(4x^2-x+6)^3$