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To solve the problem and find \(\frac{dy}{dx}\), where \(-y = \sin^{-3}(x)\), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[
-y = \sin^{-3}(x)
\]
which can be rewritten as:
\[
y = -\sin^{-3}(x)
\]

### Step 2: Differentiate both sides with respect to \(x\)
We need to differentiate \(y\) with respect to \(x\). The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\).

\[
\frac{dy}{dx} = \frac{d}{dx}\left[-\sin^{-3}(x)\right]
\]

### Step 3: Apply the chain rule
The chain rule is applied here because we are differentiating a composite function. Let \( u = \sin(x) \), then:

\[
y = -u^{-3}
\]

Now, differentiate \( y = -u^{-3} \) with respect to \( u \) and then multiply by the derivative of \( u \) with respect to \( x \):

\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
\]

First, differentiate \( -u^{-3} \) with respect to \( u \):

\[
\frac{dy}{du} = 3u^{-4}
\]

Next, differentiate \( \sin(x) \) with respect to \( x \):

\[
\frac{du}{dx} = \cos(x)
\]

So, combining these using the chain rule:

\[
\frac{dy}{dx} = 3\sin^{-4}(x) \cdot \cos(x)
\]

### Final Answer:
\[
\frac{dy}{dx} = 3\sin^{-4}(x) \cdot \cos(x)
\]

Would you like to know more details about this problem, or have any further questions?

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Here are five related questions you might be interested in:

1. How does the chain rule work in more complex functions?
2. What are some common mistakes to avoid when differentiating trigonometric functions?
3. How do we differentiate inverse trigonometric functions?
4. What are the real-world applications of trigonometric derivatives?
5. Can you explain the significance of the power rule in differentiation?

**Tip:** Always double-check if you need to apply the product rule, chain rule, or any other differentiation technique to avoid common errors.

Find \(\frac{dy}{dx}\) if \(-y = \sin^{-3}(x)\).

Find the derivative of the function -y = sin^{-3}(x) with respect to x. This problem involves the application of the chain rule and differentiation of trigonometric functions, suitable for students in Grades 10-12.

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