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

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

Use chain rule: outer derivative of \sin(u) is \cos(u), and u=x^2. So \frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x=2x\cos(x^2).

Q: Why does a factor 2x appear?

Because u=x^2. Then \frac{du}{dx}=\frac{d}{dx}(x^2)=2x. Chain rule multiplies \cos(x^2) by \frac{du}{dx}, giving 2x\cos(x^2).

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

Chain rule again. \frac{d}{dx}\cos(u)=-\sin(u)\frac{du}{dx}. With u=x^2, \frac{d}{dx}\cos(x^2)=-\sin(x^2)\cdot 2x=-2x\sin(x^2).

Q: What is \frac{d}{dx}\sin(5x^2) ?

Let u=5x^2. Then \frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}. Here \frac{du}{dx}=10x. So result is 10x\cos(5x^2).

Q: What is the derivative of \sin(x^2)+x ?

Differentiate term by term. \frac{d}{dx}\sin(x^2)=2x\cos(x^2) and \frac{d}{dx}x=1. Combine: 2x\cos(x^2)+1.

Q: What is \frac{d}{dx}\sin(x^2) evaluated at x=1?

Substitute into 2x\cos(x^2). At x=1, \frac{d}{dx}\sin(1^2)=2(1)\cos(1)=2\cos(1).

Answer

Use the chain rule. Let \(u = x^2\). Then \(\frac{d}{dx}\sin(u) = \cos(u)\cdot \frac{du}{dx}\), and \(\frac{du}{dx} = 2x\).

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

Detailed Explanation

We want the derivative of \( \sin(x^2) \).

Step 1: Identify the outer and inner functions.

The function is a composition of two functions:

Outer function: \( \sin(u) \)
Inner function: \( u = x^2 \)

Step 2: Differentiate using the Chain Rule.

The Chain Rule says:

\[
\frac{d}{dx}\bigl(\sin(u)\bigr)=\cos(u)\cdot \frac{du}{dx}.
\]

Step 3: Differentiate the inner function \(u=x^2\).

\[
u=x^2 \quad \Rightarrow \quad \frac{du}{dx}=2x.
\]

Step 4: Substitute back into the Chain Rule formula.

Now replace \(u\) with \(x^2\):

\[
\frac{d}{dx}\bigl(\sin(x^2)\bigr)=\cos(x^2)\cdot 2x.
\]

Final Answer:

\[
\frac{d}{dx}\bigl(\sin(x^2)\bigr)=2x\cos(x^2).
\]

See full solution

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Calculus FAQ

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

Use chain rule: outer derivative of \(\sin(u)\) is \(\cos(u)\), and \(u=x^2\). So \(\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x=2x\cos(x^2)\).

Why does a factor \(2x\) appear?

Because \(u=x^2\). Then \(\frac{du}{dx}=\frac{d}{dx}(x^2)=2x\). Chain rule multiplies \(\cos(x^2)\) by \(\frac{du}{dx}\), giving \(2x\cos(x^2)\).

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

Chain rule again. \(\frac{d}{dx}\cos(u)=-\sin(u)\frac{du}{dx}\). With \(u=x^2\), \(\frac{d}{dx}\cos(x^2)=-\sin(x^2)\cdot 2x=-2x\sin(x^2)\).

What is \( \frac{d}{dx}\sin(5x^2) \)?

Let \(u=5x^2\). Then \(\frac{d}{dx}\sin(u)=\cos(u)\frac{du}{dx}\). Here \(\frac{du}{dx}=10x\). So result is \(10x\cos(5x^2)\).

What is the derivative of \( \sin(x^2)+x \)?

Differentiate term by term. \(\frac{d}{dx}\sin(x^2)=2x\cos(x^2)\) and \(\frac{d}{dx}x=1\). Combine: \(2x\cos(x^2)+1\).

What is \( \frac{d}{dx}\sin(x^2)\) evaluated at \(x=1\)?

Substitute into \(2x\cos(x^2)\). At \(x=1\), \(\frac{d}{dx}\sin(1^2)=2(1)\cos(1)=2\cos(1)\).

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