Math / Calculus

Question(1 point)
Differentiate $y=\arctan (\cos x)$ . $y^{\prime}=$ $\square$ Preview My Answers Submit Answers

Studdy Solution

STEP 1

What is this asking? Find the derivative of a function that's the arctangent of cosine of $x$ . Watch out! Don't mix up the derivatives of $\arctan(x)$ and $\cos(x)$ .
Also, remember the chain rule!

STEP 2

1. Recall the derivative of arctangent
2. Recall the derivative of cosine
3. Apply the chain rule

STEP 3

The derivative of $\arctan(x)$ is $\frac{1}{1 + x^2}$ .
Remember this!
It's a key piece of the puzzle.

STEP 4

The derivative of $\cos(x)$ is $-\sin(x)$ .
Another important piece!

STEP 5

Our function $y$ is a composition of two functions: $\arctan$ and $\cos$ .
The chain rule says we take the derivative of the outer function, leaving the inner function untouched, and then multiply by the derivative of the inner function.

STEP 6

Let's apply the chain rule!
The outer function is $\arctan$ , and its derivative is $\frac{1}{1 + (\text{inner function})^2}$ .
The inner function is $\cos(x)$ , so we get $\frac{1}{1 + (\cos(x))^2}$ .

STEP 7

Now, multiply by the derivative of the inner function, which is $-\sin(x)$ .

STEP 8

Putting it all together, we get: $y' = \frac{1}{1 + (\cos(x))^2} \cdot (-\sin(x))$

STEP 9

Let's simplify that a bit! $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$

STEP 10

The derivative is $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$ .
Boom!

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Question(1 point)
Differentiate $y=\arctan (\cos x)$ . $y^{\prime}=$ $\square$ Preview My Answers Submit Answers

Studdy Solution

STEP 1

What is this asking? Find the derivative of a function that's the arctangent of cosine of $x$ . Watch out! Don't mix up the derivatives of $\arctan(x)$ and $\cos(x)$ .
Also, remember the chain rule!

STEP 2

1. Recall the derivative of arctangent
2. Recall the derivative of cosine
3. Apply the chain rule

STEP 3

The derivative of $\arctan(x)$ is $\frac{1}{1 + x^2}$ .
Remember this!
It's a key piece of the puzzle.

STEP 4

The derivative of $\cos(x)$ is $-\sin(x)$ .
Another important piece!

STEP 5

Our function $y$ is a composition of two functions: $\arctan$ and $\cos$ .
The chain rule says we take the derivative of the outer function, leaving the inner function untouched, and then multiply by the derivative of the inner function.

STEP 6

Let's apply the chain rule!
The outer function is $\arctan$ , and its derivative is $\frac{1}{1 + (\text{inner function})^2}$ .
The inner function is $\cos(x)$ , so we get $\frac{1}{1 + (\cos(x))^2}$ .

STEP 7

Now, multiply by the derivative of the inner function, which is $-\sin(x)$ .

STEP 8

Putting it all together, we get: $y' = \frac{1}{1 + (\cos(x))^2} \cdot (-\sin(x))$

STEP 9

Let's simplify that a bit! $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$

STEP 10

The derivative is $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$ .
Boom!

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Question(1 point)Differentiate y=arctan⁡(cos⁡x)y=\arctan (\cos x)y=arctan(cosx). y′=y^{\prime}=y′= □\square□ Preview My Answers Submit Answers

Studdy Solution

STEP 1

What is this asking? Find the derivative of a function that's the arctangent of cosine of xxx. Watch out! Don't mix up the derivatives of arctan⁡(x)\arctan(x)arctan(x) and cos⁡(x)\cos(x)cos(x).Also, remember the chain rule!

STEP 2

1. Recall the derivative of arctangent2. Recall the derivative of cosine3. Apply the chain rule

STEP 3

The derivative of arctan⁡(x)\arctan(x)arctan(x) is 11+x2\frac{1}{1 + x^2}1+x21​.Remember this!It's a key piece of the puzzle.

STEP 4

The derivative of cos⁡(x)\cos(x)cos(x) is −sin⁡(x)-\sin(x)−sin(x).Another important piece!

STEP 5

Our function yyy is a composition of two functions: arctan⁡\arctanarctan and cos⁡\coscos.The **chain rule** says we take the derivative of the outer function, leaving the inner function untouched, and then multiply by the derivative of the inner function.

STEP 6

Let's **apply the chain rule**!The outer function is arctan⁡\arctanarctan, and its derivative is 11+(inner function)2\frac{1}{1 + (\text{inner function})^2}1+(inner function)21​.The inner function is cos⁡(x)\cos(x)cos(x), so we get 11+(cos⁡(x))2\frac{1}{1 + (\cos(x))^2}1+(cos(x))21​.

STEP 7

Now, multiply by the derivative of the inner function, which is −sin⁡(x)-\sin(x)−sin(x).

STEP 8

Putting it all together, we get: y′=11+(cos⁡(x))2⋅(−sin⁡(x)) y' = \frac{1}{1 + (\cos(x))^2} \cdot (-\sin(x)) y′=1+(cos(x))21​⋅(−sin(x))

STEP 9

Let's **simplify** that a bit! y′=−sin⁡(x)1+cos⁡2(x) y' = \frac{-\sin(x)}{1 + \cos^2(x)} y′=1+cos2(x)−sin(x)​

STEP 10

The derivative is y′=−sin⁡(x)1+cos⁡2(x)y' = \frac{-\sin(x)}{1 + \cos^2(x)}y′=1+cos2(x)−sin(x)​.Boom!

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Question(1 point)
Differentiate $y=\arctan (\cos x)$ . $y^{\prime}=$ $\square$ Preview My Answers Submit Answers

What is this asking? Find the derivative of a function that's the arctangent of cosine of $x$ . Watch out! Don't mix up the derivatives of $\arctan(x)$ and $\cos(x)$ .
Also, remember the chain rule!

1. Recall the derivative of arctangent
2. Recall the derivative of cosine
3. Apply the chain rule

The derivative of $\arctan(x)$ is $\frac{1}{1 + x^2}$ .
Remember this!
It's a key piece of the puzzle.

The derivative of $\cos(x)$ is $-\sin(x)$ .
Another important piece!

Our function $y$ is a composition of two functions: $\arctan$ and $\cos$ .
The chain rule says we take the derivative of the outer function, leaving the inner function untouched, and then multiply by the derivative of the inner function.

Let's apply the chain rule!
The outer function is $\arctan$ , and its derivative is $\frac{1}{1 + (\text{inner function})^2}$ .
The inner function is $\cos(x)$ , so we get $\frac{1}{1 + (\cos(x))^2}$ .

Now, multiply by the derivative of the inner function, which is $-\sin(x)$ .

Putting it all together, we get: $y' = \frac{1}{1 + (\cos(x))^2} \cdot (-\sin(x))$

Let's simplify that a bit! $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$

The derivative is $y' = \frac{-\sin(x)}{1 + \cos^2(x)}$ .
Boom!