Math / Calculus

QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} g(x)=\frac{1}{\csc ^{2}(9 x)} \\ g^{\prime}(x)=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that has a trigonometric function inside, which is inside another function! Watch out! Remember the chain rule, and don't mix up the trigonometric functions and their derivatives.
Also, be careful with the signs!

STEP 2

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

STEP 3

Let's rewrite the function $g(x)$ to make it easier to differentiate.
Remember that $\frac{1}{\csc^2(9x)}$ is the same as $\sin^2(9x)$ .
So, we have: $g(x) = \sin^2(9x)$ This makes it much clearer what we're working with!

STEP 4

Now, we'll apply the chain rule.
Think of $g(x)$ as a composition of two functions: the outer function is squaring something, and the inner function is the sine of $9x$ .

STEP 5

The chain rule says that the derivative of a composition of functions is the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
Let's do it step by step!

STEP 6

The derivative of the outer function $u^2$ with respect to $u$ is $2u$ .
In our case, $u$ is $\sin(9x)$ , so the derivative of the outer function evaluated at the inner function is $2\sin(9x)$ .

STEP 7

Next, the derivative of the inner function $\sin(9x)$ with respect to $x$ is $\cos(9x) \cdot 9$ (remember to multiply by the derivative of $9x$ which is just $9$ because of the chain rule again!).

STEP 8

Putting it all together, we multiply the results from the previous two sub-steps: $g'(x) = 2\sin(9x) \cdot \cos(9x) \cdot 9$

STEP 9

We can simplify this a bit.
We know that $2\sin(a)\cos(a) = \sin(2a)$ .
In our case, $a$ is $9x$ .
So, we have: $g'(x) = 9 \cdot 2\sin(9x)\cos(9x) = 9\sin(18x)$ Look at that, nice and clean!

STEP 10

The derivative of $g(x) = \frac{1}{\csc^2(9x)}$ is $g'(x) = 9\sin(18x)$ .

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QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} g(x)=\frac{1}{\csc ^{2}(9 x)} \\ g^{\prime}(x)=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that has a trigonometric function inside, which is inside another function! Watch out! Remember the chain rule, and don't mix up the trigonometric functions and their derivatives.
Also, be careful with the signs!

STEP 2

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

STEP 3

Let's rewrite the function $g(x)$ to make it easier to differentiate.
Remember that $\frac{1}{\csc^2(9x)}$ is the same as $\sin^2(9x)$ .
So, we have: $g(x) = \sin^2(9x)$ This makes it much clearer what we're working with!

STEP 4

Now, we'll apply the chain rule.
Think of $g(x)$ as a composition of two functions: the outer function is squaring something, and the inner function is the sine of $9x$ .

STEP 5

The chain rule says that the derivative of a composition of functions is the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
Let's do it step by step!

STEP 6

The derivative of the outer function $u^2$ with respect to $u$ is $2u$ .
In our case, $u$ is $\sin(9x)$ , so the derivative of the outer function evaluated at the inner function is $2\sin(9x)$ .

STEP 7

Next, the derivative of the inner function $\sin(9x)$ with respect to $x$ is $\cos(9x) \cdot 9$ (remember to multiply by the derivative of $9x$ which is just $9$ because of the chain rule again!).

STEP 8

Putting it all together, we multiply the results from the previous two sub-steps: $g'(x) = 2\sin(9x) \cdot \cos(9x) \cdot 9$

STEP 9

We can simplify this a bit.
We know that $2\sin(a)\cos(a) = \sin(2a)$ .
In our case, $a$ is $9x$ .
So, we have: $g'(x) = 9 \cdot 2\sin(9x)\cos(9x) = 9\sin(18x)$ Look at that, nice and clean!

STEP 10

The derivative of $g(x) = \frac{1}{\csc^2(9x)}$ is $g'(x) = 9\sin(18x)$ .

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QuestionFind the derivative with respect to the independent variable. g(x)=1csc⁡2(9x)g′(x)=□\begin{array}{l} g(x)=\frac{1}{\csc ^{2}(9 x)} \\ g^{\prime}(x)=\square \end{array}g(x)=csc2(9x)1​g′(x)=□​

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that has a trigonometric function inside, which is inside another function! Watch out! Remember the chain rule, and don't mix up the trigonometric functions and their derivatives.Also, be careful with the signs!

STEP 2

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

STEP 3

Let's **rewrite** the function g(x)g(x)g(x) to make it easier to differentiate.Remember that 1csc⁡2(9x)\frac{1}{\csc^2(9x)}csc2(9x)1​ is the same as sin⁡2(9x)\sin^2(9x)sin2(9x).So, we have: g(x)=sin⁡2(9x)g(x) = \sin^2(9x)g(x)=sin2(9x) This makes it much clearer what we're working with!

STEP 4

Now, we'll **apply the chain rule**.Think of g(x)g(x)g(x) as a composition of two functions: the **outer function** is squaring something, and the **inner function** is the sine of 9x9x9x.

STEP 5

The chain rule says that the derivative of a composition of functions is the derivative of the outer function (evaluated at the inner function) *times* the derivative of the inner function.Let's do it step by step!

STEP 6

The derivative of the **outer function** u2u^2u2 with respect to uuu is 2u2u2u.In our case, uuu is sin⁡(9x)\sin(9x)sin(9x), so the derivative of the outer function evaluated at the inner function is 2sin⁡(9x)2\sin(9x)2sin(9x).

STEP 7

Next, the derivative of the **inner function** sin⁡(9x)\sin(9x)sin(9x) with respect to xxx is cos⁡(9x)⋅9\cos(9x) \cdot 9cos(9x)⋅9 (remember to multiply by the derivative of 9x9x9x which is just 999 because of the chain rule again!).

STEP 8

Putting it all together, we **multiply** the results from the previous two sub-steps: g′(x)=2sin⁡(9x)⋅cos⁡(9x)⋅9g'(x) = 2\sin(9x) \cdot \cos(9x) \cdot 9g′(x)=2sin(9x)⋅cos(9x)⋅9

STEP 9

STEP 10

The derivative of g(x)=1csc⁡2(9x)g(x) = \frac{1}{\csc^2(9x)}g(x)=csc2(9x)1​ is g′(x)=9sin⁡(18x)g'(x) = 9\sin(18x)g′(x)=9sin(18x).

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QuestionFind the derivative with respect to the independent variable. $\begin{array}{l} g(x)=\frac{1}{\csc ^{2}(9 x)} \\ g^{\prime}(x)=\square \end{array}$

What is this asking? We need to find the derivative of a function that has a trigonometric function inside, which is inside another function! Watch out! Remember the chain rule, and don't mix up the trigonometric functions and their derivatives.
Also, be careful with the signs!

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

Let's rewrite the function $g(x)$ to make it easier to differentiate.
Remember that $\frac{1}{\csc^2(9x)}$ is the same as $\sin^2(9x)$ .
So, we have: $g(x) = \sin^2(9x)$ This makes it much clearer what we're working with!

Now, we'll apply the chain rule.
Think of $g(x)$ as a composition of two functions: the outer function is squaring something, and the inner function is the sine of $9x$ .

The chain rule says that the derivative of a composition of functions is the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
Let's do it step by step!

The derivative of the outer function $u^2$ with respect to $u$ is $2u$ .
In our case, $u$ is $\sin(9x)$ , so the derivative of the outer function evaluated at the inner function is $2\sin(9x)$ .

Next, the derivative of the inner function $\sin(9x)$ with respect to $x$ is $\cos(9x) \cdot 9$ (remember to multiply by the derivative of $9x$ which is just $9$ because of the chain rule again!).

Putting it all together, we multiply the results from the previous two sub-steps: $g'(x) = 2\sin(9x) \cdot \cos(9x) \cdot 9$

The derivative of $g(x) = \frac{1}{\csc^2(9x)}$ is $g'(x) = 9\sin(18x)$ .