Math / Calculus

QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} f(x)=\frac{\cos x^{9}}{\cos ^{2} x} \\ f^{\prime}(x)=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that's a fraction with cosine functions, where the top has $x$ raised to the ninth power inside the cosine and the bottom has cosine squared. Watch out! Don't forget the chain rule, and remember how to differentiate cosine and powers of cosine!
Also, don't mix up the quotient rule and the product rule.

STEP 2

1. Rewrite the function
2. Apply the quotient rule
3. Simplify the derivative

STEP 3

Let's rewrite our function $f(x)$ to make it easier to take the derivative.
We can rewrite $\frac{1}{\cos^2 x}$ as $(\cos x)^{-2}$ .
So, $f(x) = \frac{\cos x^9}{\cos^2 x}$ becomes $f(x) = (\cos x^9) \cdot (\cos x)^{-2}$ .
This will make applying the product rule easier!

STEP 4

Remember the quotient rule: If we have a function $f(x) = \frac{g(x)}{h(x)}$ , then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ .
Here, $g(x) = \cos(x^9)$ and $h(x) = (\cos x)^2$ .

STEP 5

Let's find $g'(x)$ .
We have $g(x) = \cos(x^9)$ .
Using the chain rule, $g'(x) = -\sin(x^9) \cdot 9x^8 = -9x^8\sin(x^9)$ .

STEP 6

Now let's find $h'(x)$ .
We have $h(x) = (\cos x)^2$ .
Using the chain rule, $h'(x) = 2(\cos x)(-\sin x) = -2\sin x \cos x$ .

STEP 7

Putting it all together with the quotient rule: $f'(x) = \frac{(-9x^8\sin(x^9))(\cos^2 x) - (\cos(x^9))(-2\sin x \cos x)}{(\cos^2 x)^2}$

STEP 8

Let's simplify our derivative: $f'(x) = \frac{-9x^8\sin(x^9)\cos^2 x + 2\sin x \cos x \cos(x^9)}{\cos^4 x}$

STEP 9

We can factor out $\cos x$ from the numerator: $f'(x) = \frac{\cos x(-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9))}{\cos^4 x}$

STEP 10

Now, we can divide both the numerator and denominator by $\cos x$ : $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$

STEP 11

The derivative of $f(x) = \frac{\cos x^9}{\cos^2 x}$ is $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$ .

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QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} f(x)=\frac{\cos x^{9}}{\cos ^{2} x} \\ f^{\prime}(x)=\square \end{array}$

Studdy Solution

STEP 1

STEP 2

1. Rewrite the function
2. Apply the quotient rule
3. Simplify the derivative

STEP 3

Let's rewrite our function $f(x)$ to make it easier to take the derivative.
We can rewrite $\frac{1}{\cos^2 x}$ as $(\cos x)^{-2}$ .
So, $f(x) = \frac{\cos x^9}{\cos^2 x}$ becomes $f(x) = (\cos x^9) \cdot (\cos x)^{-2}$ .
This will make applying the product rule easier!

STEP 4

Remember the quotient rule: If we have a function $f(x) = \frac{g(x)}{h(x)}$ , then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ .
Here, $g(x) = \cos(x^9)$ and $h(x) = (\cos x)^2$ .

STEP 5

Let's find $g'(x)$ .
We have $g(x) = \cos(x^9)$ .
Using the chain rule, $g'(x) = -\sin(x^9) \cdot 9x^8 = -9x^8\sin(x^9)$ .

STEP 6

Now let's find $h'(x)$ .
We have $h(x) = (\cos x)^2$ .
Using the chain rule, $h'(x) = 2(\cos x)(-\sin x) = -2\sin x \cos x$ .

STEP 7

Putting it all together with the quotient rule: $f'(x) = \frac{(-9x^8\sin(x^9))(\cos^2 x) - (\cos(x^9))(-2\sin x \cos x)}{(\cos^2 x)^2}$

STEP 8

Let's simplify our derivative: $f'(x) = \frac{-9x^8\sin(x^9)\cos^2 x + 2\sin x \cos x \cos(x^9)}{\cos^4 x}$

STEP 9

We can factor out $\cos x$ from the numerator: $f'(x) = \frac{\cos x(-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9))}{\cos^4 x}$

STEP 10

Now, we can divide both the numerator and denominator by $\cos x$ : $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$

STEP 11

The derivative of $f(x) = \frac{\cos x^9}{\cos^2 x}$ is $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$ .

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QuestionFind the derivative with respect to the independent variable. f(x)=cos⁡x9cos⁡2xf′(x)=□\begin{array}{l} f(x)=\frac{\cos x^{9}}{\cos ^{2} x} \\ f^{\prime}(x)=\square \end{array}f(x)=cos2xcosx9​f′(x)=□​

Studdy Solution

STEP 1

STEP 2

1. Rewrite the function2. Apply the quotient rule3. Simplify the derivative

STEP 3

STEP 4

STEP 5

Let's find g′(x)g'(x)g′(x).We have g(x)=cos⁡(x9)g(x) = \cos(x^9)g(x)=cos(x9).Using the **chain rule**, g′(x)=−sin⁡(x9)⋅9x8=−9x8sin⁡(x9)g'(x) = -\sin(x^9) \cdot 9x^8 = -9x^8\sin(x^9)g′(x)=−sin(x9)⋅9x8=−9x8sin(x9).

STEP 6

Now let's find h′(x)h'(x)h′(x).We have h(x)=(cos⁡x)2h(x) = (\cos x)^2h(x)=(cosx)2.Using the **chain rule**, h′(x)=2(cos⁡x)(−sin⁡x)=−2sin⁡xcos⁡xh'(x) = 2(\cos x)(-\sin x) = -2\sin x \cos xh′(x)=2(cosx)(−sinx)=−2sinxcosx.

STEP 7

Putting it all together with the **quotient rule**: f′(x)=(−9x8sin⁡(x9))(cos⁡2x)−(cos⁡(x9))(−2sin⁡xcos⁡x)(cos⁡2x)2f'(x) = \frac{(-9x^8\sin(x^9))(\cos^2 x) - (\cos(x^9))(-2\sin x \cos x)}{(\cos^2 x)^2}f′(x)=(cos2x)2(−9x8sin(x9))(cos2x)−(cos(x9))(−2sinxcosx)​

STEP 8

Let's **simplify** our derivative: f′(x)=−9x8sin⁡(x9)cos⁡2x+2sin⁡xcos⁡xcos⁡(x9)cos⁡4xf'(x) = \frac{-9x^8\sin(x^9)\cos^2 x + 2\sin x \cos x \cos(x^9)}{\cos^4 x}f′(x)=cos4x−9x8sin(x9)cos2x+2sinxcosxcos(x9)​

STEP 9

We can **factor out** cos⁡x\cos xcosx from the numerator: f′(x)=cos⁡x(−9x8sin⁡(x9)cos⁡x+2sin⁡xcos⁡(x9))cos⁡4xf'(x) = \frac{\cos x(-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9))}{\cos^4 x}f′(x)=cos4xcosx(−9x8sin(x9)cosx+2sinxcos(x9))​

STEP 10

Now, we can **divide both** the numerator and denominator by cos⁡x\cos xcosx: f′(x)=−9x8sin⁡(x9)cos⁡x+2sin⁡xcos⁡(x9)cos⁡3xf'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}f′(x)=cos3x−9x8sin(x9)cosx+2sinxcos(x9)​

STEP 11

The derivative of f(x)=cos⁡x9cos⁡2xf(x) = \frac{\cos x^9}{\cos^2 x}f(x)=cos2xcosx9​ is f′(x)=−9x8sin⁡(x9)cos⁡x+2sin⁡xcos⁡(x9)cos⁡3xf'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}f′(x)=cos3x−9x8sin(x9)cosx+2sinxcos(x9)​.

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QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} f(x)=\frac{\cos x^{9}}{\cos ^{2} x} \\ f^{\prime}(x)=\square \end{array}$

1. Rewrite the function
2. Apply the quotient rule
3. Simplify the derivative

Let's find $g'(x)$ .
We have $g(x) = \cos(x^9)$ .
Using the chain rule, $g'(x) = -\sin(x^9) \cdot 9x^8 = -9x^8\sin(x^9)$ .

Now let's find $h'(x)$ .
We have $h(x) = (\cos x)^2$ .
Using the chain rule, $h'(x) = 2(\cos x)(-\sin x) = -2\sin x \cos x$ .

Putting it all together with the quotient rule: $f'(x) = \frac{(-9x^8\sin(x^9))(\cos^2 x) - (\cos(x^9))(-2\sin x \cos x)}{(\cos^2 x)^2}$

Let's simplify our derivative: $f'(x) = \frac{-9x^8\sin(x^9)\cos^2 x + 2\sin x \cos x \cos(x^9)}{\cos^4 x}$

We can factor out $\cos x$ from the numerator: $f'(x) = \frac{\cos x(-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9))}{\cos^4 x}$

Now, we can divide both the numerator and denominator by $\cos x$ : $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$

The derivative of $f(x) = \frac{\cos x^9}{\cos^2 x}$ is $f'(x) = \frac{-9x^8\sin(x^9)\cos x + 2\sin x \cos(x^9)}{\cos^3 x}$ .