Q. \[ \frac{d}{dx}\left(\tan^2(x)\right) \]

Q: What is \frac{d}{dx}\big(\tan^2(x)\big) ?

Use chain rule. \frac{d}{dx}(\tan^2 x)=2\tan x\cdot \sec^2 x .

Q: Can you show the chain rule steps for \tan^2(x) ?

Let u=\tan x. Then \frac{d}{dx}(u^2)=2u\frac{du}{dx}=2\tan x\cdot \sec^2 x .

Q: Is the derivative equivalently 2\tan^3(x)+2\tan(x) ?

Yes. Since \sec^2 x=1+\tan^2 x, then 2\tan x\sec^2 x=2\tan x(1+\tan^2 x)=2\tan x+2\tan^3 x.

Q: What is \frac{d}{dx}(\tan^2(3x)) ?

Chain rule again. \frac{d}{dx}\big(\tan^2(3x)\big)=2\tan(3x)\sec^2(3x)\cdot 3=6\tan(3x)\sec^2(3x) .

Answer

Use the chain rule. Let \(y=\tan^2(x)\). Then

\[
\frac{dy}{dx}=2\tan(x)\cdot \frac{d}{dx}\big(\tan(x)\big)
\]
\[
\frac{d}{dx}\big(\tan(x)\big)=\sec^2(x)
\]

So the derivative is

\[
\frac{d}{dx}\big(\tan^2(x)\big)=2\tan(x)\sec^2(x)
\]

Detailed Explanation

We want to find the derivative of the function

\[
f(x)=\tan^2(x).
\]

Step 1: Identify the outer and inner functions (chain rule idea).

Notice that \( \tan^2(x) \) can be viewed as an outer function applied to an inner function:

\[
f(x)=\left(\tan(x)\right)^2.
\]

Step 2: Differentiate the outer function.

If we let \(u=\tan(x)\), then \(f(x)=u^2\). The derivative of \(u^2\) with respect to \(u\) is

\[
\frac{d}{du}\left(u^2\right)=2u.
\]

Step 3: Differentiate the inner function.

Now we need the derivative of \(u=\tan(x)\) with respect to \(x\). We use the known derivative

\[
\frac{d}{dx}\left(\tan(x)\right)=\sec^2(x).
\]

Step 4: Apply the chain rule.

The chain rule says

\[
\frac{d}{dx}\left(u^2\right)=2u\cdot \frac{du}{dx}.
\]

Substitute back \(u=\tan(x)\) and \(\frac{du}{dx}=\sec^2(x)\):

\[
f'(x)=2\tan(x)\cdot \sec^2(x).
\]

Final Answer.

\[
\frac{d}{dx}\left(\tan^2(x)\right)=2\tan(x)\sec^2(x).
\]

See full solution

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Calculus FAQ

What is \( \frac{d}{dx}\big(\tan^2(x)\big) \)?

Use chain rule. \( \frac{d}{dx}(\tan^2 x)=2\tan x\cdot \sec^2 x \).

Can you show the chain rule steps for \( \tan^2(x) \)?

Let \(u=\tan x\). Then \( \frac{d}{dx}(u^2)=2u\frac{du}{dx}=2\tan x\cdot \sec^2 x \).

Is the derivative equivalently \(2\tan^3(x)+2\tan(x)\) ?

Yes. Since \( \sec^2 x=1+\tan^2 x\), then \(2\tan x\sec^2 x=2\tan x(1+\tan^2 x)=2\tan x+2\tan^3 x\).

What is \( \frac{d}{dx}(\tan^2(3x)) \)?

Chain rule again. \( \frac{d}{dx}\big(\tan^2(3x)\big)=2\tan(3x)\sec^2(3x)\cdot 3=6\tan(3x)\sec^2(3x) \).

What is \( \frac{d}{dx}(\sec^2(x)) \) for comparison?

\( \frac{d}{dx}(\sec^2 x)=2\sec x\cdot (\sec x\tan x)=2\sec^2 x\tan x \).

Why can’t we treat \(\tan^2(x)\) as \((\tan(x))^2\) without chain rule?

Actually we do treat it that way; the extra part comes from differentiating the inside \( \tan(x) \). The chain rule is required because \( \tan(x) \) depends on \(x\).

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