Find D Z D T \frac{d Z}{d T} D T D Z ​ Where Z = ( X + 2 Y ) 10 Z=(x+2 Y)^{10} Z = ( X + 2 Y ) 10 , X = Sin ⁡ T X=\sin T X = Sin T , And Y = 4 T 3 Y=4 T^3 Y = 4 T 3 .

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Introduction

Implicit differentiation is a powerful technique used to find the derivative of a function that is not easily expressed in terms of a single variable. In this article, we will explore how to use implicit differentiation to find $\frac{d z}{d t}$ where $z=(x+2 y)^{10}$, $x=\sin t$, and $y=4 t^3$.

Understanding the Problem

We are given a function $z=(x+2 y)^{10}$, where $x$ and $y$ are both functions of $t$. Our goal is to find the derivative of $z$ with respect to $t$, denoted as $\frac{d z}{d t}$. To do this, we will use the chain rule and implicit differentiation.

Step 1: Differentiate the Outer Function

The outer function is $u^{10}$, where $u=x+2 y$. To differentiate this function with respect to $t$, we will use the chain rule. The chain rule states that if we have a composite function $f(g(t))$, then the derivative of this function with respect to $t$ is given by $f'(g(t)) \cdot g'(t)$.

In this case, we have $f(u)=u^{10}$ and $g(t)=x+2 y$. To find the derivative of $f(u)$ with respect to $u$, we will use the power rule, which states that if $f(u)=u^n$, then $f'(u)=n u^{n-1}$.

Applying the power rule, we get:

$$\frac{d f}{d u} = 10 u^9$$

Step 2: Differentiate the Inner Function

The inner function is $g(t)=x+2 y$. To differentiate this function with respect to $t$, we will use the sum rule and the chain rule.

The sum rule states that if we have a function $f(x)+g(x)$, then the derivative of this function with respect to $x$ is given by $f'(x)+g'(x)$.

In this case, we have $g(t)=x+2 y$. To find the derivative of $g(t)$ with respect to $t$, we will use the chain rule.

We know that $x=\sin t$ and $y=4 t^3$. To find the derivative of $x$ with respect to $t$, we will use the chain rule. The derivative of $\sin t$ with respect to $t$ is given by $\cos t$.

To find the derivative of $y$ with respect to $t$, we will use the power rule. The derivative of $4 t^3$ with respect to $t$ is given by $12 t^2$.

Applying the chain rule, we get:

$$\frac{d g}{d t} = \frac{d x}{d t} + 2 \frac{d y}{d t}$$

$$\frac{d g}{d t} = \cos t + 2 (12 t^2)$$

$$\frac{d g}{d t} = \cos t + 24 t^2$$

Step 3: Apply the Chain Rule

Now that we have found the derivatives of the outer and inner functions, we can apply the chain rule to find the derivative of $z$ with respect to $t$.

The chain rule states that if we have a composite function $f(g(t))$, then the derivative of this function with respect to $t$ is given by $f'(g(t)) \cdot g'(t)$.

In this case, we have $f(u)=u^{10}$ and $g(t)=x+2 y$. To find the derivative of $z$ with respect to $t$, we will multiply the derivatives of the outer and inner functions.

Applying the chain rule, we get:

$$\frac{d z}{d t} = \frac{d f}{d u} \cdot \frac{d g}{d t}$$

$$\frac{d z}{d t} = 10 u^9 \cdot (\cos t + 24 t^2)$$

Step 4: Substitute the Value of $u$

We know that $u=x+2 y$. To find the value of $u$, we will substitute the values of $x$ and $y$ into this equation.

We know that $x=\sin t$ and $y=4 t^3$. Substituting these values into the equation for $u$, we get:

$$u = \sin t + 2 (4 t^3)$$

$$u = \sin t + 8 t^3$$

Step 5: Substitute the Value of $u$ into the Equation for $\frac{d z}{d t}$

Now that we have found the value of $u$, we can substitute this value into the equation for $\frac{d z}{d t}$.

We know that $\frac{d z}{d t} = 10 u^9 \cdot (\cos t + 24 t^2)$. Substituting the value of $u$ into this equation, we get:

$$\frac{d z}{d t} = 10 (\sin t + 8 t^3)^9 \cdot (\cos t + 24 t^2)$$

Conclusion

In this article, we used implicit differentiation to find $\frac{d z}{d t}$ where $z=(x+2 y)^{10}$, $x=\sin t$, and $y=4 t^3$. We applied the chain rule and the power rule to find the derivative of the outer and inner functions, and then multiplied these derivatives to find the derivative of $z$ with respect to $t$.

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Introduction

Implicit differentiation is a powerful technique used to find the derivative of a function that is not easily expressed in terms of a single variable. In this article, we will explore some common questions and answers related to implicit differentiation.

Q: What is implicit differentiation?

A: Implicit differentiation is a technique used to find the derivative of a function that is not easily expressed in terms of a single variable. It involves differentiating both sides of an equation with respect to the variable, while treating the other variables as constants.

Q: When should I use implicit differentiation?

A: You should use implicit differentiation when you have a function that is not easily expressed in terms of a single variable. This can occur when you have a function that involves multiple variables, or when you have a function that is defined implicitly.

Q: How do I apply implicit differentiation?

A: To apply implicit differentiation, you will need to follow these steps:

1. Differentiate both sides of the equation with respect to the variable.
2. Treat the other variables as constants.
3. Use the chain rule and the power rule to find the derivative of the function.

Q: What is the chain rule?

A: The chain rule is a technique used to find the derivative of a composite function. It states that if you have a composite function $f(g(t))$, then the derivative of this function with respect to $t$ is given by $f'(g(t)) \cdot g'(t)$.

Q: What is the power rule?

A: The power rule is a technique used to find the derivative of a function that involves a power. It states that if you have a function $f(x)=x^n$, then the derivative of this function with respect to $x$ is given by $f'(x)=n x^{n-1}$.

Q: How do I find the derivative of a function that involves multiple variables?

A: To find the derivative of a function that involves multiple variables, you will need to use the chain rule and the power rule. You will also need to treat the other variables as constants.

Q: What are some common mistakes to avoid when using implicit differentiation?

A: Some common mistakes to avoid when using implicit differentiation include:

• Failing to treat the other variables as constants.
• Failing to use the chain rule and the power rule.
• Failing to check your work.

Q: How do I check my work when using implicit differentiation?

A: To check your work when using implicit differentiation, you will need to follow these steps:

1. Check that you have differentiated both sides of the equation correctly.
2. Check that you have treated the other variables as constants.
3. Check that you have used the chain rule and the power rule correctly.

Conclusion

Implicit differentiation is a powerful technique used to find the derivative of a function that is not easily expressed in terms of a single variable. By following the steps outlined in this article, you can use implicit differentiation to find the derivative of a wide range of functions.

Common Applications of Implicit Differentiation

Implicit differentiation has a wide range of applications in mathematics and science. Some common applications include:

• Finding the derivative of a function that involves multiple variables.
• Finding the derivative of a function that is defined implicitly.
• Finding the derivative of a function that involves a power.

Real-World Examples of Implicit Differentiation

Implicit differentiation has many real-world applications. Some examples include:

• Finding the rate of change of a quantity that is defined implicitly.
• Finding the derivative of a function that is used to model a real-world phenomenon.
• Finding the derivative of a function that is used to solve a real-world problem.

Conclusion

Implicit differentiation is a powerful technique used to find the derivative of a function that is not easily expressed in terms of a single variable. By following the steps outlined in this article, you can use implicit differentiation to find the derivative of a wide range of functions.