How do you differentiate y= (3+2^x)^xy=(3+2x)x?
1 Answer
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:
To deal with
\ \ \ \ \ 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))