Determine (dy/dx) using implicit differentiation

As such, the above solution should not be used as your own solution. Instead, use it as a guide to help you solve the problem on your own. Remember, the goal is to learn and understand the material, not just get the answer.
  • #1
koolkris623
19
0
Determine (dy/dx) using implicit differentiation.

cos(X^2Y^2) = x

I'm really confused what to do now..i think the next steps are:

d/dx [cos(X^2*Y^2)] = d/dx [x]
= -sin(X^2*Y^2)* ((X^2*2Y dy/dx) + (Y^2*2X)) = 1
= -2YX^2 sin(X^2*Y^2) dy/dx + -2XY^2sin(X^2*Y^2) = 1
= -2YX^2 sin(X^2*Y^2) dy/dx = 1 + -2XY^2sin(X^2*Y^2)
= dy/dx = (1 + -2XY^2sin(X^2*Y^2))/ (-2YX^2 sin(X^2*Y^2))

Can someone tell me if this is correct?
 
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  • #2
[tex]cos(x^2 y^2)=x[/tex]

[tex]\frac{d}{dx}[cos(x^2 y^2)]=\frac{d}{dx}[x][/tex]

[tex]-sin(x^2 y^2) [\frac{d}{dx}(x^2 y^2)] = 1[/tex]

[tex]-sin(x^2 y^2) [2y^2x + 2x^2y\frac{dy}{dx}] = 1[/tex]

[tex]-sin(x^2 y^2)2y^2x -sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1[/tex]

[tex]-sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1+sin(x^2 y^2)2y^2x [/tex]

[tex]\frac{dy}{dx} = \frac{1+sin(x^2 y^2)2y^2x}{-sin(x^2 y^2)2x^2y} [/tex]
 
  • #3
cool thanks
 
  • #4
atqamar, please note that one should not provide full solutions in the homework forums, but rather should offer guidance and hints as to how to proceed.
 

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly defined in terms of one variable. This means that the variable y is not written as a function of x, but rather as an equation involving both x and y. It is often used when it is difficult or impossible to solve for y in terms of x.

How do you use implicit differentiation to find (dy/dx)?

To find the derivative using implicit differentiation, we treat y as a function of x and use the chain rule to differentiate both sides of the equation. This means that we take the derivative of each term with respect to x and then multiply by the derivative of the inside function, which in this case is y'. The end result will be an expression for (dy/dx).

What is the purpose of implicit differentiation?

The purpose of implicit differentiation is to find the derivative of a function that is not explicitly defined in terms of one variable. This is useful in situations where it is difficult or impossible to express one variable in terms of the other, but we still need to find the rate of change of the function.

What are some common applications of implicit differentiation?

Implicit differentiation is commonly used in physics and engineering to find the rate of change of quantities that are related by an equation, such as velocity and acceleration. It is also used in economics to find the marginal rate of substitution between two goods, and in biology to find the rate of change of a population in a predator-prey system.

Are there any limitations to implicit differentiation?

Implicit differentiation can only be used to find the derivative of a function that is not explicitly defined in terms of one variable. Additionally, it can be a more complicated and time-consuming method compared to other techniques, so it may not be the most efficient option in some cases. Lastly, implicit differentiation does not provide an explicit formula for (dy/dx), so it may not be suitable for finding exact values.

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