What is the derivative of #5^tanx#?

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Answer 1 · 2018-03-03T10:48:00

#=> (dy)/(dx) = 5^tanx * ln5 * sec^2 x #

We are trying to find #(dy)/(dx) # when #y = 5^tanx #

Taking natural logs on both sides...

#=> ln y = ln 5^tanx #

Using log laws:

#alpha log beta -= log beta ^ alpha #

#=> ln y =ln5 * tanx #

Differentiating implicitly:

#=> 1/y * (dy)/(dx) = ln5 * sec^2 x #

#=> (dy)/(dx) = y ln5 * sec^2 x #

#=> (dy)/(dx) = 5^tanx * ln5 * sec^2 x #