Q. \[ \text{Find the derivative of } \cos^{2}(x). \]

Q: What is the derivative of \cos^2(x) ?\n

Use the chain rule: \dfrac{d}{dx}\left(\cos^2(x)\right)=2\cos(x)\cdot(-\sin(x))=-2\sin(x)\cos(x).\n\n

Q: How do I differentiate \left(\cos(x)\right)^2 by the chain rule ?\n

Let u=\cos(x). Then \dfrac{d}{dx}(u^2)=2u\cdot\dfrac{du}{dx}. Since \dfrac{du}{dx}=-\sin(x), the result is -2\sin(x)\cos(x).\n\n

Q: Can I rewrite \cos^2(x) using a trig identity first ?\n

Yes: \cos^2(x)=\dfrac{1+\cos(2x)}{2}. Then \dfrac{d}{dx}\cos^2(x)=\dfrac{1}{2}\cdot\dfrac{d}{dx}\cos(2x)=\dfrac{1}{2}\cdot(-2\sin(2x))=-\sin(2x).\n\n

Q: Are -2\sin(x)\cos(x) and -\sin(2x) equivalent ?\n

Yes. Use \sin(2x)=2\sin(x)\cos(x). Then -\sin(2x)=-2\sin(x)\cos(x), so both forms match.\n\n

Q: What is the derivative of f(x)=\cos^2(3x) ?\n

Let u=3x. Then \dfrac{d}{dx}\cos^2(u)=2\cos(u)\cdot(-\sin(u))\cdot\dfrac{du}{dx}=-2\cos(3x)\sin(3x)\cdot 3=-6\sin(3x)\cos(3x).\n\n

Q: What is the derivative of \cos^2(x) using product form ?\n

Write \cos^2(x)=\cos(x)\cos(x). Then f'(x)=\cos'(x)\cos(x)+\cos(x)\cos'(x)=(-\sin(x))\cos(x)+\cos(x)(-\sin(x))=-2\sin(x)\cos(x).

Answer

Let \(y=\cos^2(x)\). Use the chain rule:

\[
\frac{dy}{dx}=2\cos(x)\cdot\frac{d}{dx}(\cos(x))=2\cos(x)\cdot(-\sin(x))
\]

\[
\frac{dy}{dx}=-2\cos(x)\sin(x)
\]

Final result: \(-2\cos(x)\sin(x)\).

Detailed Explanation

We want to find the derivative of \( \cos^2(x) \). Here \( \cos^2(x) \) means \(\big(\cos(x)\big)^2\), not \(\cos(x^2)\).

Step 1: Identify the outer and inner functions

We can rewrite the function as a composition:

\[
f(x)=\big(\cos x\big)^2
\]

Let

\[
u(x)=\cos x
\]

Then

\[
f(x)=\big(u(x)\big)^2
\]

Step 2: Use the chain rule

The chain rule says:

\[
\frac{d}{dx}\big(u(x)^2\big)=2u(x)\cdot u'(x)
\]

Step 3: Differentiate the inner function

Since \(u(x)=\cos x\), we have:

\[
u'(x)=\frac{d}{dx}(\cos x)=-\sin x
\]

Step 4: Substitute back into the chain rule

Substitute \(u(x)=\cos x\) and \(u'(x)=-\sin x\):

\[
\frac{d}{dx}\big(\cos x\big)^2=2(\cos x)(-\sin x)
\]

Step 5: Simplify

Multiply the constants and signs:

\[
2(\cos x)(-\sin x)=-2\sin x\cos x
\]

Final Answer

\[
\boxed{\frac{d}{dx}\left(\cos^2(x)\right)=-2\sin(x)\cos(x)}
\]

See full solution

Graph

Need help with derivatives? Try our AI homework tools today!

AI Homework Helper

Calculus FAQ

What is the derivative of \(\cos^2(x)\) ?\n

Use the chain rule: \(\dfrac{d}{dx}\left(\cos^2(x)\right)=2\cos(x)\cdot(-\sin(x))=-2\sin(x)\cos(x)\).\n\n

How do I differentiate \(\left(\cos(x)\right)^2\) by the chain rule ?\n

Let \(u=\cos(x)\). Then \(\dfrac{d}{dx}(u^2)=2u\cdot\dfrac{du}{dx}\). Since \(\dfrac{du}{dx}=-\sin(x)\), the result is \(-2\sin(x)\cos(x)\).\n\n

Can I rewrite \(\cos^2(x)\) using a trig identity first ?\n

Yes: \(\cos^2(x)=\dfrac{1+\cos(2x)}{2}\). Then \(\dfrac{d}{dx}\cos^2(x)=\dfrac{1}{2}\cdot\dfrac{d}{dx}\cos(2x)=\dfrac{1}{2}\cdot(-2\sin(2x))=-\sin(2x)\).\n\n

Are \(-2\sin(x)\cos(x)\) and \(-\sin(2x)\) equivalent ?\n

Yes. Use \(\sin(2x)=2\sin(x)\cos(x)\). Then \(-\sin(2x)=-2\sin(x)\cos(x)\), so both forms match.\n\n

What is the derivative of \(f(x)=\cos^2(3x)\) ?\n

Let \(u=3x\). Then \(\dfrac{d}{dx}\cos^2(u)=2\cos(u)\cdot(-\sin(u))\cdot\dfrac{du}{dx}=-2\cos(3x)\sin(3x)\cdot 3=-6\sin(3x)\cos(3x)\).\n\n

What is the derivative of \(\cos^2(x)\) using product form ?\n

Write \(\cos^2(x)=\cos(x)\cos(x)\). Then \(f'(x)=\cos'(x)\cos(x)+\cos(x)\cos'(x)=(-\sin(x))\cos(x)+\cos(x)(-\sin(x))=-2\sin(x)\cos(x)\).

Use these AI tools for practice.
Get help for derivate problems.

298,376+ active customers
Math, Geometry, Trigonometry, etc.

AI Math Solver Geometry AI Trig Solver