Find D Y D T \frac{dy}{dt} D T D Y ​ For The Function Y = ( 1 + Sin ⁡ 9 T ) − 5 Y=(1+\sin 9t)^{-5} Y = ( 1 + Sin 9 T ) − 5 .A) − 45 ( Cos ⁡ 9 T ) − 6 -45(\cos 9t)^{-6} − 45 ( Cos 9 T ) − 6 B) − 5 ( 1 + Sin ⁡ 9 T ) − 6 -5(1+\sin 9t)^{-6} − 5 ( 1 + Sin 9 T ) − 6 C) − 45 ( 1 + Sin ⁡ 9 T ) − 6 Cos ⁡ 9 T -45(1+\sin 9t)^{-6} \cos 9t − 45 ( 1 + Sin 9 T ) − 6 Cos 9 T D) − 5 ( 1 + Sin ⁡ 9 T ) − 6 Cos ⁡ 9 T -5(1+\sin 9t)^{-6} \cos 9t − 5 ( 1 + Sin 9 T ) − 6 Cos 9 T

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Introduction

In this article, we will explore the concept of finding the derivative of a function with a trigonometric expression. The given function is $y=(1+\sin 9t)^{-5}$, and we are asked to find $\frac{dy}{dt}$. This problem requires the application of the chain rule and the knowledge of trigonometric derivatives.

The Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. The chain rule states that if we have a function of the form $y=f(g(x))$, then the derivative of $y$ with respect to $x$ is given by $\frac{dy}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}$.

Applying the Chain Rule

In this case, we have a function of the form $y=(1+\sin 9t)^{-5}$. We can rewrite this function as $y=f(g(t))$, where $f(u)=u^{-5}$ and $g(t)=1+\sin 9t$. Using the chain rule, we can find the derivative of $y$ with respect to $t$ as follows:

$$\frac{dy}{dt}=\frac{df}{dg}\cdot\frac{dg}{dt}$$

Finding $\frac{df}{dg}$

To find $\frac{df}{dg}$, we need to find the derivative of $f(u)$ with respect to $u$. Using the power rule, we have:

$$\frac{df}{du}=-5u^{-6}$$

Finding $\frac{dg}{dt}$

To find $\frac{dg}{dt}$, we need to find the derivative of $g(t)$ with respect to $t$. Using the chain rule, we have:

$$\frac{dg}{dt}=\frac{d}{dt}(1+\sin 9t)=9\cos 9t$$

Combining the Results

Now that we have found $\frac{df}{du}$ and $\frac{dg}{dt}$, we can combine the results to find $\frac{dy}{dt}$:

$$\frac{dy}{dt}=\frac{df}{du}\cdot\frac{dg}{dt}=-5(1+\sin 9t)^{-6}\cdot9\cos 9t$$

Simplifying the Result

We can simplify the result by combining the constants:

$$\frac{dy}{dt}=-45(1+\sin 9t)^{-6}\cos 9t$$

Conclusion

In this article, we have found the derivative of the function $y=(1+\sin 9t)^{-5}$ with respect to $t$. The result is $\frac{dy}{dt}=-45(1+\sin 9t)^{-6}\cos 9t$. This problem requires the application of the chain rule and the knowledge of trigonometric derivatives.

Final Answer

The final answer is $\boxed{-45(1+\sin 9t)^{-6}\cos 9t}$.

Discussion

This problem is a classic example of the chain rule in action. The chain rule allows us to find the derivative of a composite function by breaking it down into smaller components. In this case, we have a function of the form $y=(1+\sin 9t)^{-5}$, and we need to find $\frac{dy}{dt}$. The chain rule allows us to find the derivative of this function by finding the derivative of the inner function $g(t)=1+\sin 9t$ and then multiplying it by the derivative of the outer function $f(u)=u^{-5}$.

Related Problems

• Find the derivative of the function $y=(1+\cos 2t)^{-3}$ with respect to $t$.
• Find the derivative of the function $y=(2+\sin 3t)^{-2}$ with respect to $t$.
• Find the derivative of the function $y=(3-\cos 4t)^{-4}$ with respect to $t$.

Solutions to Related Problems

• Find the derivative of the function $y=(1+\cos 2t)^{-3}$ with respect to $t$. The derivative of this function is $\frac{dy}{dt}=-3(1+\cos 2t)^{-4}\cdot(-2\sin 2t)=6(1+\cos 2t)^{-4}\sin 2t$.
• Find the derivative of the function $y=(2+\sin 3t)^{-2}$ with respect to $t$. The derivative of this function is $\frac{dy}{dt}=-2(2+\sin 3t)^{-3}\cdot3\cos 3t=-6(2+\sin 3t)^{-3}\cos 3t$.
• Find the derivative of the function $y=(3-\cos 4t)^{-4}$ with respect to $t$. The derivative of this function is $\frac{dy}{dt}=-4(3-\cos 4t)^{-5}\cdot4\sin 4t=-16(3-\cos 4t)^{-5}\sin 4t$.

Conclusion

In this article, we have found the derivative of the function $y=(1+\sin 9t)^{-5}$ with respect to $t$. The result is $\frac{dy}{dt}=-45(1+\sin 9t)^{-6}\cos 9t$. This problem requires the application of the chain rule and the knowledge of trigonometric derivatives. We have also provided solutions to related problems and discussed the chain rule in action.

Introduction

In our previous article, we explored the concept of finding the derivative of a function with a trigonometric expression using the chain rule. In this article, we will provide a Q&A section to help clarify any doubts and provide additional examples.

Q: What is the chain rule?

A: The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. It states that if we have a function of the form $y=f(g(x))$, then the derivative of $y$ with respect to $x$ is given by $\frac{dy}{dx}=\frac{df}{dg}\cdot\frac{dg}{dx}$.

Q: How do I apply the chain rule?

A: To apply the chain rule, you need to identify the inner function $g(x)$ and the outer function $f(u)$. Then, you need to find the derivative of the inner function $\frac{dg}{dx}$ and the derivative of the outer function $\frac{df}{du}$. Finally, you multiply the two derivatives together to get the final result.

Q: What if I have a function with multiple layers of composition?

A: If you have a function with multiple layers of composition, you can apply the chain rule multiple times. For example, if you have a function of the form $y=f(g(h(x)))$, you can first find the derivative of the innermost function $h(x)$, then the derivative of the next innermost function $g(h(x))$, and finally the derivative of the outermost function $f(g(h(x)))$.

Q: Can I use the chain rule with other types of functions?

A: Yes, the chain rule can be used with other types of functions, such as exponential functions, logarithmic functions, and trigonometric functions. However, you need to be careful when applying the chain rule with these types of functions, as the derivatives may be more complex.

Q: How do I find the derivative of a function with a trigonometric expression?

A: To find the derivative of a function with a trigonometric expression, you can use the chain rule and the knowledge of trigonometric derivatives. For example, if you have a function of the form $y=(1+\sin x)^{-5}$, you can first find the derivative of the inner function $g(x)=1+\sin x$, then the derivative of the outer function $f(u)=u^{-5}$, and finally multiply the two derivatives together.

Q: What are some common mistakes to avoid when applying the chain rule?

A: Some common mistakes to avoid when applying the chain rule include:

• Forgetting to identify the inner function and the outer function
• Forgetting to find the derivative of the inner function
• Forgetting to find the derivative of the outer function
• Not multiplying the two derivatives together
• Not simplifying the final result

Q: Can I use the chain rule with functions that have multiple variables?

A: Yes, the chain rule can be used with functions that have multiple variables. However, you need to be careful when applying the chain rule with multiple variables, as the derivatives may be more complex.

Q: How do I find the derivative of a function with multiple variables using the chain rule?

A: To find the derivative of a function with multiple variables using the chain rule, you can first identify the inner functions and the outer functions, then find the derivatives of the inner functions and the outer functions, and finally multiply the derivatives together.

Q: What are some examples of functions that can be differentiated using the chain rule?

A: Some examples of functions that can be differentiated using the chain rule include:

• $y=(1+\sin x)^{-5}$
• $y=(2+\cos 2x)^{-3}$
• $y=(3-\sin 3x)^{-4}$
• $y=(1+\cos 4x)^{-2}$
• $y=(2-\sin 5x)^{-5}$

Conclusion

In this article, we have provided a Q&A section to help clarify any doubts and provide additional examples of finding derivatives with the chain rule. We have also discussed common mistakes to avoid and provided examples of functions that can be differentiated using the chain rule.