How do you find the derivative of a natural log exponential function #y = (ln(x^2))^(2x+3)#?

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Answer 1 · 2015-08-23T21:28:22

When we see the variable in both the base and the exponent, we need some form of logarithmic differentiation to find the derivative.

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

So

#lny = ln((lnx^2)^(2x+3))#

#lny = (2x+3) ln(lnx^2)#

Differentiate implicitly.

On the left we get: #1/y dy/dx#

On the right use the product rule, the derivative of #ln# and the chain rule to get some expression #G(x)#

We'll have: #1/y dy/dx = G(x)#

so #dy/dx = y G(x)#

# = (lnx^2)^(2x+3) G(x)#