How do you find #dy/dx# by implicit differentiation given #e^x=lny#?

Answer 1

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

Use the following differentiation rules:

•#d/dx (e^x) = e^x#
•#d/dx (lnx) = 1/x#
When we use implicit differentiation, we must differentiate with respect to one variable without isolating another.  
The following step represents that we take the derivative with respect to #x# on each side of the relation.
#d/dx(e^x) = d/dx(ln y)#

Use the identities above, and implicit differentiation:

#e^x = 1/y(dy/dx)#
#e^x/(1/y) = dy/dx#
#dy/dx = ye^x#
The relation #e^x = lny# is simple enough such that we can find an explicit equation for #y# to resubstitute into the derivative.
#e^x = lny#
#e^(e^x) = y#
We could have also differentiated the above equation to find #dy/dx#, but it's simpler just using implicit differentiation. Since #y= e^(e^x)#
#dy/dx = e^(e^x)e^x#
Use the rule #(a^n)(a^m) = a^(n + m)#:
#dy/dx = e^(e^x + x)#

Hopefully this helps!

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Answer 2

To find ( \frac{dy}{dx} ) by implicit differentiation given ( e^x = \ln y ), differentiate both sides of the equation with respect to ( x ), then solve for ( \frac{dy}{dx} ).

Differentiate both sides with respect to ( x ):

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

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

Solve for ( \frac{dy}{dx} ):

[ \frac{dy}{dx} = e^x \cdot y ]

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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.

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.

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.

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.

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