Find dy du , du dx , and dy dx .

(a) y = u5 and u = x2 + 1 dy du = du dx = dy dx =

(b) y = u4 and u = 4x2 − x + 6 dy du = du dx = dy dx =

Respuesta :

Answer:

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

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

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 [tex]\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}[/tex].

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

[tex]\frac{du}{dx}=\frac{d}{dx} (x^2 +1)=2x[/tex].

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

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

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

Then, [tex]\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[/tex]  

Answer:

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Step-by-step explanation:

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