Find the derivative of the given function.f(x)=cot−1(ecosx)f\left(x\right)=\cot^{-1}\left(e^{\cos x}\right)f(x)=cot−1(ecosx)

sin ⁡ x ⋅ e cos ⁡ x 1 + e 2 cos ⁡ x \frac{\sin x\cdot e^{\cos x}}{1+e^{2\cos x}} 1 + e 2 c o s x s i n x ⋅ e c o s x

Find the derivative of the given function. f ( x ) = cot − 1 ( e cos x ) f\left(x\right)=\cot^{-1}\left(e^{\cos x}\right) f ( x ) = cot − 1 ( e c o s x )

Hey, everyone. Here, we're asked to take the derivative of this function, f of x, is equal to the inverse cotangent of e to the power of the cosine of x. Now you'll only be able to take this derivative if you know all of your rules for derivatives. So if you have not yet learned how to find the derivative of e to the power of something, then you'll wanna circle back to this problem later. For the rest of you, we know how to find this derivative, f prime of x, and we can see here that we have a function inside of a function inside of another function. So that means we're gonna have to apply the chain rule here multiple times, continuing to multiply by each successive derivative, working our way from the outside to the inside. So starting with our outermost function here, that's the inverse cotangent. This is going to be negative one over 1 plus whatever's inside that cotangent. So here, e to the power of the cosine of x, that is squared. Then working my way inside, this is e to the power of the cosine of x. This derivative is just e to the power of the cosine of x. Multiplying by the derivative of that innermost function, the derivative of cosine is negative sine. So this last derivative, negative sine of x. Now we can clean this up a little bit, just rewriting this a little bit more compactly. These two negatives will cancel out, so I'm left with a positive. So this is e to the power of the cosine of x times the sine of x over 1+e to the power of 2cosine of x, just condensing that a little bit more. So this is my full derivative here. Now we applied a bunch of different rules for derivatives throughout this process, and you are doing a great job at getting through all of them. Thanks for watching, and I'll see you in the next one.

Find the derivative of the given function. f ( x ) = cot ⁡ − 1 ( e cos ⁡ x ) f\left(x\right)=\cot^{-1}\left(e^{\cos x}\right) f ( x ) = cot − 1 ( e c o s x )

sin ⁡ x ⋅ e cos ⁡ x 1 + e 2 cos ⁡ x \frac{\sin x\cdot e^{\cos x}}{1+e^{2\cos x}} 1 + e 2 c o s x s i n x ⋅ e c o s x

− 1 1 + e 2 cos ⁡ x -\frac{1}{1+e^{2\cos x}} − 1 + e 2 c o s x 1

− e cos ⁡ x 1 + e 2 cos ⁡ x -\frac{e^{\cos x}}{1+e^{2\cos x}} − 1 + e 2 c o s x e c o s x

sin ⁡ x ⋅ e cos ⁡ x 1 + x 2 \frac{\sin x\cdot e^{\cos x}}{1+x^2} 1 + x 2 s i n x ⋅ e c o s x

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the derivative of the given function.
$f\left(x\right)=\cot^{-1}\left(e^{\cos x}\right)$

$-\frac{1}{1+e^{2\cos x}}$

$-\frac{e^{\cos x}}{1+e^{2\cos x}}$

$\frac{\sin x\cdot e^{\cos x}}{1+e^{2\cos x}}$

$\frac{\sin x\cdot e^{\cos x}}{1+x^2}$

Verified step by step guidance

Identify the function to differentiate: $ f(x) = \cot^{-1}(e^{\cos x}) $. This is a composition of functions, so we will use the chain rule.

Recall the derivative of $ \cot^{-1}(u) $ with respect to $ u $ is $ -\frac{1}{1+u^2} $. Here, $ u = e^{\cos x} $.

Apply the chain rule: $ \frac{d}{dx}[\cot^{-1}(e^{\cos x})] = \frac{d}{du}[\cot^{-1}(u)] \cdot \frac{du}{dx} $.

Calculate $ \frac{du}{dx} $ where $ u = e^{\cos x} $. Use the chain rule again: $ \frac{d}{dx}[e^{\cos x}] = e^{\cos x} \cdot (-\sin x) $.

Combine the results: $ \frac{d}{dx}[\cot^{-1}(e^{\cos x})] = -\frac{1}{1+(e^{\cos x})^2} \cdot e^{\cos x} \cdot (-\sin x) $. Simplify to get the final expression.

7:26m

Watch next

Master Derivatives of Inverse Sine & Inverse Cosine with a bite sized video explanation from Patrick

Start learning

Derivatives of Inverse Trigonometric Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice