2.8 Implicit Differentiation

Transcription

1 .8 Implicit Differentiation Section.8 Notes Page 1 Before I tell ou what implicit differentiation is, let s start with an example: EXAMPLE: Find if x. This question is asking us to find the derivative of with respect to x. In other words, x has to be the onl variable. We can definitel solve for. To do this, factor out a. You will get ( x 1). Then divide to 1 get:. To get the derivative of this one we can first rewrite this as: ( x 1). Now we can appl x 1 the power rule to get: ( x 1) (1). This can be rewritten as:. ( x 1) EXAMPLE: Find if x. The problem with this one is that if we tr and solve for we won t end up with an expression that has onl x in it. Therefore we must do this b implicit differentiation. We will take the derivative of both sides of this equation without tring to solve for. Whenever I take the derivative of an x term, the power rule can be applied as alwas. What about the terms? Since we don t know what is whenever we get to a part where we need to take the derivative of a, we will get. For a term like, when we take the derivative we will be appling the chain rule. The outside function can be done b using the power rule. The inside function is, so again the derivative of this will be. The derivative of with respect to x will be 1. Here is what the whole thing will look like once I take the derivative of both sides: x 0. Notice there was a single term next to the equals sign. The derivative of is. You want to now solve for since that is what the question is asking us to find. x All the terms that have ( x 1) We factored out the will be left on one side of the equation.. Now divide both sides b - + 1, which is also 1. x This is our answer. Usuall ou will have both x and in our answer. 1 Now that we have seen the process of implicit differentiation, let s redo the first example we did onl now we will use implicit differentiation.

2 Section.8 Notes Page EXAMPLE: Find if x. This time we will not solve for. To do implicit we must again take the derivative of both sides of the equation. On left side of the equation we have x. This is not a single term. It is two terms multiplied together. For this one we must use the product rule. When we get to part of the product rule where we take the derivative of we will alwas get. Here is what the problem looks like after taking the derivative of both sides: f g g f x ( 1) 0 Now we need to get all terms with on one side of the equation. x Now factor out the. ( x 1) Now divide both sides b x + 1 to get the answer:. x 1 You will notice this is not the same answer as our example. To get that answer let s look at our first example. in that problem we got. If we put this in for we will get: x 1. After simplifing we will x 1 x 1 get: which is the same answer as in the first example. For all the problems in this section we ( x 1) will not know what is so we don t need to substitute in anthing for. I just did it here to show ou that implicit will give ou the correct formula for the derivative. EXAMPLE: Use implicit differentiation to find if x x. When we take the derivative of both sides we need to use the product rule twice on the left hand side because there are two separate terms where x and are multiplied together. When ou use the product rule, when it gets to the part where we just write g, we are not taking the derivative of, so that is wh we don t put a in this term. This happens with the derivative of. x term below: The onl time ou have a is when ou actuall take the f g g f f g g f x x (1) x 0 Get all the terms on one side of the equation. x x x Now factor out the.

3 Section.8 Notes Page ( x x) x Divide both sides b x x. x Last step is to factor. x x (x ) This is our answer. x( x ) EXAMPLE: Use implicit differentiation to find if ( x ) sin. We will take the derivative of both sides. On the left side the chain rule is applied. First the power rule is used to bring down. Then I did the derivative of the inside: (x ) x (1) 0 cos Note that for the derivative of x, I needed to appl the product rule. Note the derivative of is f g g f x () On the right side I need to take the derivative of sin. This one also involves the chain rule. First I took the derivative of sine, which is cosine. But then I needed to take the derivative of, which is where the came from. After simplifing the original result, we get: x 6 x cos. We need to multipl the two parenthesis together: 8x 8x 1x 1 cos. Now we will get all the terms on one side of the equation 8x 1x cos 8x 1. Factor out. 8 x 1x cos 8x 1 Solve for. 8x 1 8x 1x cos This is our answer. More on next page

4 Section.8 Notes Page EXAMPLE: Solve for b using implicit differentiation: e x x 10. For the first term, u x and u x (1). So we need to take the derivative of both sides: e x x x x x 0. We can multipl the first two terms: xe e x 0. Now we need to get all the terms with on one side of the equation: x e x xe x. Now factor out a from the left hand side: x x x xe e x e x. Solve for to get:. x xe d EXAMPLE: Use implicit differentiation to find and if x 8. x 0 Here we used the power rule on to do the derivative of the outside function. Once again the inside function is, so that is wh there is an extra attached to that term. x You want all the terms with on one side of the equation. x x B dividing both sides b x. we get Now we need to find the second derivative. We need to take the derivative again and use the quotient rule: ( x) ( x ) d I will simplif this expression. x x d I will divide both things on top b. d x x x We will substitute for since we found this earlier. d x x x Simplif. d x x. Now we can get common denominators. d x x This is our final answer for the second derivative.

5 Section.8 Notes Page EXAMPLE: Verif that (, 1) is on the curve x x 1. Then find lines that are (a) tangent and (b) normal to the curve at (, 1). To verif if the point is on the curve, plug in a for x and a 1 for : 1 ()(1) 1. If ou simplif both sides ou will get 9 9. Since both sides are the same it verifies that (, 1) is on the curve. Now for part (a). (a) We must first find the derivative b implicit differentiation. On the right hand side we will use the product rule: f g g f x x We want all the terms with on one side of the equation.. x x Now factor out the ( x x) Now divide both sides b x. x This is our derivative. Now put in our point. So x = and = 1. x (1) () So we now know our slope, m, is. We will use = mx + b to find b. (1) () () b, and after solving for b we get b. The line 8 11 x is tangent to the curve at (, 1). (b) If a line is normal to the curve, this means ou want a line that is perpendicular to the tangent line but still passes through (, 1). The line that is perpendicular will have a slope of / 8. We will use = mx + b to find b: 1 () b, and after solving for b we get b 9/. Our equation is /8x 9/. 8 EXAMPLE: Verif that 1, 0 is on the curve cos( x ) x. Then find lines that are (a) tangent and (b) normal to the curve at 1, 0. To verif if the point is on the curve, plug in a 1 for x and 0 for : cos( (1) 0) 0 (1 ). If ou simplif both sides ou will get cos. So ( 1) the curve.. Since both sides are the same it verifies that, 1 0 is on (a) We must first find the derivative b implicit differentiation. On the left hand side we will use the chain rule. The derivative of the outside will give us sin( x ). Then we need to multipl this b the derivative of the inside, which is x. The derivative is. After the cosine term we had a single so its derivative is just. After the equals sign we have a x, so its derivative is. Putting this all together gives us:

6 Section.8 Notes Page 6 sin( x ) First distribute to get individual terms. sin( x ) sin( x ) We will keep terms with on one side of equation. sin( x ) sin( x ) Now factor out the. Divide both sides b what is after the. sin( x ) 1 sin( x ) sin( x ) sin( x ) 1 sin( (1) 0) sin sin( (1) 0) 1 sin 1 Now we can plug in our point 0, 1. So we know that m since sin 0. We will use = mx + b to find b. tangent to the curve at (1, 0). 0 (1) b, and after solving for b we get b. The line x is (b) The line that is normal will have a slope of 1/. We will use = mx + b to find b. So 0 (1/ )(1) b, and after solving for b we get b 1/. The line normal to the curve at (1, 0) is ( 1/ ) x 1/. EX AMPLE: You are given that x cos 1. Find b implicit differentiation and evaluate the derivative at the given point,,. The left side of the equation is a product, so we will first need to use the product rule. f g g f x ( sin ) cos (1) 0 Now solve for. xsin cos We want all the terms with to sta on one side of the equation. cos So we have that xsin cos which is the same as xsin cot. x We now want to evaluate the derivative at,. To do this, plug in a for x and a for. You will get: cot 1 1 cot. 6