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How do you do implicit differentiation to find dy/dx (Calculus)? To skip ahead: 1) For a BASIC example using the POWER RULE, skip to time 3:57. 2) For an example that uses a TRIG FUNCTION and the PRODUCT RULE, skip to time 7:20.

What is implicit differentiation? Up until now, most functions you've had to differentiate have probably been written EXPLICITLY as a function of x, such as y = x^2. However, if a function y is written IMPLICITLY as a function of x, such as x^2 + y^2 = 9, you will need to use implicit differentiation to find the derivative dy/dx. The only difference is that any time you take the derivative of y, you must also multiply by dy/dx. The reason is that y is dependent on x. You can think of y as containing some x expression inside it, so

implicit differentiation is basically a special instance of the chain rule in which you must take the outside derivative but also multiply by the inside derivative, which is dy/dx in this case.

Here are the steps to doing implicit differentiation to find DY/DX:

1) TAKE THE DERIVATIVE OF BOTH SIDES, MULTIPLYING BY DY/DX every time you take the derivative of a Y: The first step is to take the derivative of both sides of the equation, with respect to x, but to attach a dy/dx if you ever take the derivative of y. For instance, to implicitly differentiate the equation x^2 + y^2 = 9, take the derivative of both sides with respect to x. On the left, the derivative of the x^2 term is just 2x. To differentiate the y^2 term, in this case first use the power rule to get 2y, and THEN, because y is dependent on x, you must multiply the term by dy/dx so that you have y^2 times dy/dx. Don't forget to differentiate the right side of the original equation as well, which was the constant 9, so the derivative is just 0. Your new, differentiated equation is then 2x plus y^2 dy/dx = 0.

2) GET DY/DX ALONE ON ONE SIDE: The third step is to solve for dy/dx, or in other words, to get dy/dx alone on one side of the equation. Next, DISTRIBUTE AND EXPAND BOTH SIDES if necessary distributing or opening up any terms that have parentheses. Then, if dy/dx appears in more than one term, get those terms together on one side and use the SIMPLIFICATION TRICK OF FACTORING out dy/dx from those terms so that it then appears only once on that side. Finally, DIVIDE OUT any factor that is currently multiplied by the dy/dx so that you are left with dy/dx alone on one side and an equation that looks like dy/dx = some other expression.

NOTE: Only multiply by dy/dx if you are taking the derivative of y. In more challenging examples like the second example in this video, you do not need to attach a dy/dx if you are just multiplying by y. For instance, when differentiating a term like xy, using the product rule gives you x times dy/dx plus y. Notice that in the second term from the product rule, it was not necessary to attach a dy/dx to the y, since you were not taking the derivative of y in that term.

How do you do implicit differentiation to find dy/dx (Calculus)? To skip ahead: 1) For a BASIC example using the POWER RULE, skip to time 3:57. 2) For an example that uses a TRIG FUNCTION and the PRODUCT RULE, skip to time 7:20.

What is implicit differentiation? Up until now, most functions you've had to differentiate have probably been written EXPLICITLY as a function of x, such as y = x^2. However, if a function y is written IMPLICITLY as a function of x, such as x^2 + y^2 = 9, you will need to use implicit differentiation to find the derivative dy/dx. The only difference is that any time you take the derivative of y, you must also multiply by dy/dx. The reason is that y is dependent on x. You can think of y as containing some x expression inside it, so

implicit differentiation is basically a special instance of the chain rule in which you must take the outside derivative but also multiply by the inside derivative, which is dy/dx in this case.

Here are the steps to doing implicit differentiation to find DY/DX:

1) TAKE THE DERIVATIVE OF BOTH SIDES, MULTIPLYING BY DY/DX every time you take the derivative of a Y: The first step is to take the derivative of both sides of the equation, with respect to x, but to attach a dy/dx if you ever take the derivative of y. For instance, to implicitly differentiate the equation x^2 + y^2 = 9, take the derivative of both sides with respect to x. On the left, the derivative of the x^2 term is just 2x. To differentiate the y^2 term, in this case first use the power rule to get 2y, and THEN, because y is dependent on x, you must multiply the term by dy/dx so that you have y^2 times dy/dx. Don't forget to differentiate the right side of the original equation as well, which was the constant 9, so the derivative is just 0. Your new, differentiated equation is then 2x plus y^2 dy/dx = 0.

2) GET DY/DX ALONE ON ONE SIDE: The third step is to solve for dy/dx, or in other words, to get dy/dx alone on one side of the equation. Next, DISTRIBUTE AND EXPAND BOTH SIDES if necessary distributing or opening up any terms that have parentheses. Then, if dy/dx appears in more than one term, get those terms together on one side and use the SIMPLIFICATION TRICK OF FACTORING out dy/dx from those terms so that it then appears only once on that side. Finally, DIVIDE OUT any factor that is currently multiplied by the dy/dx so that you are left with dy/dx alone on one side and an equation that looks like dy/dx = some other expression.

NOTE: Only multiply by dy/dx if you are taking the derivative of y. In more challenging examples like the second example in this video, you do not need to attach a dy/dx if you are just multiplying by y. For instance, when differentiating a term like xy, using the product rule gives you x times dy/dx plus y. Notice that in the second term from the product rule, it was not necessary to attach a dy/dx to the y, since you were not taking the derivative of y in that term.

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