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How do you differentiate #y= (3+2^x)^x#?

Answer 1

# dy/dx = (3+2^x)^x((ln(2)x2^x)/(3+2^x) + ln (3+2^x))#

The simplest method is to use what is known as logarithmic differentiation. We take (Natural) logarithms of both sides and the use implicit differentiation:

# y = (3+2^x)^x #

Taking logarithms we get:

# lny = ln {(3+2^x)^x} # # lny = xln (3+2^x) #

Then we can differentiate the LHS Implicitly, and the RHS using the product rule to get: # 1/ydy/dx = (x)(d/dxln (3+2^x)) + (ln (3+2^x))(1) # # :. 1/(3+2^x)^x dy/dx = xd/dxln (3+2^x) + ln (3+2^x) ... [1]#

To deal with #d/dxln (3+2^x)# we use the chain rule:

#\ \ \ \ \ d/dxln (3+2^x) = 1/(3+2^x) d/dx(3+2^x)# # :. d/dxln (3+2^x) = 1/(3+2^x) ln(2)2^x # # :. d/dxln (3+2^x) = (ln(2)2^x)/(3+2^x) #

Substituting the last result into[1] we get:

# 1/(3+2^x)^x dy/dx = (ln(2)x2^x)/(3+2^x) + ln (3+2^x)# # dy/dx = (3+2^x)^x((ln(2)x2^x)/(3+2^x) + ln (3+2^x))#

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Answer 2

Adam Adkins

To differentiate ( y = (3 + 2^x)^x ), you can use the chain rule. The derivative is ( \frac{dy}{dx} = x(3 + 2^x)^{x-1}(3\ln(2) + 2^x\ln(3 + 2^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.