Example 3If F(x, Y) = \sin \left(\frac{6x}{1+y}\right ], Calculate ∂ F ∂ X \frac{\partial F}{\partial X} ∂ X ∂ F ​ And ∂ F ∂ Y \frac{\partial F}{\partial Y} ∂ Y ∂ F ​ .SolutionUsing The Chain Rule For Functions Of One Variable, We

Introduction

In calculus, partial derivatives are used to find the rate of change of a multivariable function with respect to one of its variables, while keeping the other variables constant. In this article, we will calculate the partial derivatives of the function $f(x, y) = \sin \left(\frac{6x}{1+y}\right)$ with respect to $x$ and $y$.

The Chain Rule

To find the partial derivatives of $f(x, y)$, we will use the Chain Rule for functions of one variable. The Chain Rule states that if we have a composite function $f(g(x))$, then the derivative of $f$ with respect to $x$ is given by:

$$\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}$$

Calculating $\frac{\partial f}{\partial x}$

To find $\frac{\partial f}{\partial x}$, we will use the Chain Rule. We can rewrite $f(x, y)$ as:

$$f(x, y) = \sin \left(\frac{6x}{1+y}\right)$$

Let $u = \frac{6x}{1+y}$. Then, we can rewrite $f(x, y)$ as:

$$f(x, y) = \sin(u)$$

Now, we can use the Chain Rule to find $\frac{\partial f}{\partial x}$:

$$\frac{\partial f}{\partial x} = \frac{df}{du} \cdot \frac{du}{dx}$$

We know that $\frac{df}{du} = \cos(u)$, and $\frac{du}{dx} = \frac{6}{1+y}$. Therefore:

$$\frac{\partial f}{\partial x} = \cos(u) \cdot \frac{6}{1+y}$$

Substituting $u = \frac{6x}{1+y}$, we get:

$$\frac{\partial f}{\partial x} = \cos \left(\frac{6x}{1+y}\right) \cdot \frac{6}{1+y}$$

Calculating $\frac{\partial f}{\partial y}$

To find $\frac{\partial f}{\partial y}$, we will use the Chain Rule. We can rewrite $f(x, y)$ as:

$$f(x, y) = \sin \left(\frac{6x}{1+y}\right)$$

Let $u = \frac{6x}{1+y}$. Then, we can rewrite $f(x, y)$ as:

$$f(x, y) = \sin(u)$$

Now, we can use the Chain Rule to find $\frac{\partial f}{\partial y}$:

$$\frac{\partial f}{\partial y} = \frac{df}{du} \cdot \frac{du}{dy}$$

We know that $\frac{df}{du} = \cos(u)$, and $\frac{du}{dy} = -\frac{6x}{(1+y)^2}$. Therefore:

$$\frac{\partial f}{\partial y} = \cos(u) \cdot -\frac{6x}{(1+y)^2}$$

Substituting $u = \frac{6x}{1+y}$, we get:

$$\frac{\partial f}{\partial y} = -\cos \left(\frac{6x}{1+y}\right) \cdot \frac{6x}{(1+y)^2}$$

Conclusion

In this article, we calculated the partial derivatives of the function $f(x, y) = \sin \left(\frac{6x}{1+y}\right)$ with respect to $x$ and $y$. We used the Chain Rule for functions of one variable to find the partial derivatives. The partial derivatives are:

$$\frac{\partial f}{\partial x} = \cos \left(\frac{6x}{1+y}\right) \cdot \frac{6}{1+y}$$

$$\frac{\partial f}{\partial y} = -\cos \left(\frac{6x}{1+y}\right) \cdot \frac{6x}{(1+y)^2}$$

Introduction

In our previous article, we calculated the partial derivatives of the function $f(x, y) = \sin \left(\frac{6x}{1+y}\right)$ with respect to $x$ and $y$. In this article, we will answer some common questions related to partial derivatives.

Q: What is the difference between a partial derivative and a total derivative?

A: A partial derivative is the derivative of a function with respect to one of its variables, while keeping the other variables constant. A total derivative, on the other hand, is the derivative of a function with respect to all of its variables.

Q: How do I calculate the partial derivative of a function with respect to multiple variables?

A: To calculate the partial derivative of a function with respect to multiple variables, you can use the Chain Rule for functions of one variable. You can rewrite the function as a composite function, and then use the Chain Rule to find the partial derivative.

Q: What is the notation for partial derivatives?

A: The notation for partial derivatives is $\frac{\partial f}{\partial x}$, where $f$ is the function and $x$ is the variable with respect to which the derivative is being taken.

Q: Can I use the Power Rule to find the partial derivative of a function?

A: Yes, you can use the Power Rule to find the partial derivative of a function. The Power Rule states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. You can use this rule to find the partial derivative of a function with respect to one of its variables.

Q: How do I use the Chain Rule to find the partial derivative of a function?

A: To use the Chain Rule to find the partial derivative of a function, you can rewrite the function as a composite function. Let $u = g(x)$, where $g(x)$ is a function of $x$. Then, you can use the Chain Rule to find the partial derivative of the function with respect to $x$:

$$\frac{\partial f}{\partial x} = \frac{df}{du} \cdot \frac{du}{dx}$$

Q: Can I use the Product Rule to find the partial derivative of a function?

A: Yes, you can use the Product Rule to find the partial derivative of a function. The Product Rule states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. You can use this rule to find the partial derivative of a function with respect to one of its variables.

Q: How do I use the Quotient Rule to find the partial derivative of a function?

A: To use the Quotient Rule to find the partial derivative of a function, you can rewrite the function as a quotient of two functions. Let $f(x) = \frac{u(x)}{v(x)}$, where $u(x)$ and $v(x)$ are functions of $x$. Then, you can use the Quotient Rule to find the partial derivative of the function with respect to $x$:

$$\frac{\partial f}{\partial x} = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2}$$

Conclusion

In this article, we answered some common questions related to partial derivatives. We discussed the notation for partial derivatives, the Chain Rule, the Power Rule, the Product Rule, and the Quotient Rule. We also provided examples of how to use these rules to find the partial derivative of a function with respect to one of its variables.