How do you find the derivative of $y = ln (cos (x))$ $y=ln(cos(x))$ ?

Question

How do you find the derivative of $y = ln (cos (x))$ $y=ln(cos(x))$ ?

2 Answers

Impact of this question

163823 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-08-22T01:46:12

You can find this derivative by applying the Chain Rule, with $cos x$ $cosx$ as the inner function, and $ln x$ $lnx$ as the outer function.

Process:

To apply the chain rule, we first find the derivative of the outer function, $ln u$ $lnu$ , with $u = cos x$ $u = cosx$ . Remember that the derivative of $ln u = \frac{1}{u} = \frac{1}{cos x}$ $lnu = 1/u = 1/cosx$ .

Now we just need to find the derivative of the inner function, $cos x$ $cosx$ , and multiply it by the derivative of the outer function we just found.

Since the derivative of $cos x$ $cosx$ is ( $- sin x$ $-sinx$ ), we end up with:

$\frac{d y}{d x} = (\frac{1}{cos x}) \cdot (- sin x) = (- \frac{sin x}{cos x}) = - tan x$ $dy/dx = (1/cosx) * (-sinx) = (-sinx/cosx) = -tanx$ .

A shorter way to do these is to just know that the derivative of a $ln (u)$ $ln(u)$ -type function is the derivative of the inside over the original of what's inside.

Answer 2 · 2017-06-04T10:16:09

$\frac{d y}{d x} = - tan x$ $dy/dx=-tanx$

Explanation:

$differentiate using the chain rule$ $"differentiate using the "color(blue)"chain rule"$

$• d/dx(ln(f(x)))=(f'(x))/(f(x))$

$rArrdy/dx=(-sinx)/(cosx)=-tanx$

Science

Math

Humanities

... and beyond

How do you find the derivative of $y = ln (cos (x))$ $y=ln(cos(x))$ ?

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of y=ln(cos(x))y=ln(cos(x))y=ln(cos(x)) ?

2 Answers

Explanation:

Related questions

Impact of this question

How do you find the derivative of $y = ln (cos (x))$ $y=ln(cos(x))$ ?