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f(x) = 5^(tan5x)
We will use the chain rule to find the derivative.
Let g(x) = tan5x.
==> g'(x) = 5*sec^2 (5x)
==> f(x) = 5^g(x).
Let us differentiate using the chain rule.
==> f'(x) = [5^(g(x))' * g'(x).
= (5^(tan5x))' * (tan5x)'
= (5^tan5x * ln 5 )* (5sec^2 5x)
==> f'(x) = 5*ln 5 (5^tan5x) * sec^2 5x
But we know that sec^2 5x = 1/cos^2 5x
==> f'(x) = (5^tan5x + 1)*ln 5 / cos^2 5x.
To determine the derivative of the given composed function, we'll differentiate both sides:
dy= [5^ (tan 5x)]'dx
We'll apply chain rule:
dy = 5^ (tan 5x)*ln 5*(tan 5x)'dx
But (tan 5x)' = (5x)'/(cos 5x)^2
(tan 5x)' = 5/(cos 5x)^2
dy = (5*5^ (tan 5x)*ln 5)/(cos 5x)^2 dx
We'll add the exponents of 5:
dy/dx = (5^ (tan 5x + 1)*ln 5)/(cos 5x)^2 dx
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