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

1 Answer
Nov 29, 2016

dy/dx = (3+2^x)^x((ln(2)x2^x)/(3+2^x) + ln (3+2^x))dydx=(3+2x)x(ln(2)x2x3+2x+ln(3+2x))

Explanation:

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 y=(3+2x)x

Taking logarithms we get:

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

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) 1ydydx=(x)(ddxln(3+2x))+(ln(3+2x))(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))