How do you differentiate $f (x) = ln (tan x)$ $f(x)=ln(tanx)$ using the chain rule?

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Answer 1 · 2015-12-08T11:21:03

$\frac{1}{tan (x) \cdot {cos}^{2} (x)} = csc (x) sec (x)$ $1/(tan(x)*cos^2(x))=csc(x)sec(x)$

You need the chain rule for composite functions, and the rule states that

$d/dx f(g(x)) = f'g(x) * g'(x)$

So, you differentiate the outer function, leaving its argument untouched, and then you multiply by the derivative of the inner function.

In your case, the outer function is $f(x)=ln(x)$ , and the inner function is $g(x)=tan(x)$

So, the derivative of the outer function is $d/dx ln(x)=1/x$ , so

$f'(g(x))=1/tan(x)$ .

On the other hand, the derivative of $tan(x)$ is $1/cos^2(x)$ .

So,

$f'g(x) * g'(x) = 1/tan(x) * 1/cos^2(x) = 1/(tan(x)*cos^2(x))$

And since $tan(x)=sin(x)/cos(x)$ , we have that

$tan(x)*cos^2(x)=sin(x)cos(x)$ , and so the derivative becomes

$1/(sin(x)cos(x))=csc(x)sec(x)$

How do you differentiate f(x)=ln(tanx)f(x)=ln(tanx)f(x)=ln(tanx) using the chain rule?