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How do you find #(dy)/(dx)# given #lny=xe^x#?

Answer 1

Paisley Johnson

# y'=e^(x(e^x+1) ) (x+1)#

#lny=xe^x#

using the product rule on the RHS

#1/y \ y'=e^x + x e^x#

the rest is algebra

# y'=ye^x( 1 + x )#

# y'=e^(x e^x)e^x( 1 + x )#

# y'=e^(x(e^x+1) ) (x+1)#

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Answer 2

Samantha Adkison

#dy/dx=(x+1) e^(xe^x +x)#

#ln y = xe^x#

By implicit differentiation:

#1/y dy/dx = xe^x + e^x# (Standard differential and Product rule)

#dy/dx = (x+1) e^x * y#

But since #ln y = xe^x -> y = e^(xe^x)#

Therefore #dy/dx = (x+1) e^x * e^(xe^x)#

#dy/dx = (x+1) e^(xe^x+x)#

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Answer 3

Luke Alvarez

To find (\frac{dy}{dx}) given (\ln(y) = xe^x), differentiate both sides of the equation with respect to (x).

[ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(xe^x) ]

Using the chain rule for differentiation on the left side, and the product rule on the right side:

[ \frac{1}{y} \frac{dy}{dx} = e^x + xe^x ]

Multiply both sides by (y):

[ \frac{dy}{dx} = y(e^x + xe^x) ]

Substitute (y) back in using the original equation (\ln(y) = xe^x):

[ y = e^{xe^x} ]

[ \frac{dy}{dx} = e^{xe^x}(e^x + xe^x) ]

So, (\frac{dy}{dx} = e^x(e^x + xe^x)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.