Question

8. Find $\frac{dy}{dx}$ if $y = x^{x^2}$. [A] $\ln x + 1$ [B] $x^2 x^{x^2 - 1}$ [C] $x(2 \ln x + 1)x^{x^2}$ [D] $(\ln x)x^{x^2}$ [E] $x(\ln x - 1)x^{x^2}$

          8. Find $\frac{dy}{dx}$ if $y = x^{x^2}$.
[A] $\ln x + 1$ [B] $x^2 x^{x^2 - 1}$
[C] $x(2 \ln x + 1)x^{x^2}$ [D] $(\ln x)x^{x^2}$
[E] $x(\ln x - 1)x^{x^2}$

8. Find (dy)/(dx) if y = x^x^2.
[A] ln x + 1 [B] x^2 x^x^2 - 1
[C] x(2 ln x + 1)x^x^2 [D] (ln x)x^x^2
[E] x(ln x - 1)x^x^2

Added by Juan Jos- U.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

05/02/2022

Step 1

Let's rewrite y as y = x^u, where u = x^2. Using the chain rule, we have: (dy)/(dx) = (dy)/(du) * (du)/(dx) Show more…

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