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nadezda [96]
4 years ago
8

Find dy du , du dx , and dy dx .

Mathematics
1 answer:
dangina [55]4 years ago
8 0

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  

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