Implicit Differentiation - the basics

Transcription

1 x x 6 Implicit Differentiation - the basics Implicit differentiation is the name for the method of differentiation that we use when we have not explicitl solved for in terms of x (that means we did not solve for =... We have, instead, the mixed in with the x and/or constants). There are a couple of reasons where we might need or want to not solve for. ) Often we might have an equation where it is impossible to solve explicitl for, as in. ) Or, it might be inconvenient to solve for. A tpical equation might be that of a circle or ellipse: x 6 Even though we can solve for 6 x, taking the derivative once we have done so will be much more complicated, requiring the chain rule, risking much error, and most students forget the +/- in front of the square root and are not sure what to do with it. I see ten times as man mistakes from people who refuse to use implicit differentiation and insist on solving for in cases like this than I do from those who use the eas method, implicit differentiation. So, please, do not avoid this method - it is eas once ou get the hang of it. There are two areas which cause problems when students first start using implicit differentiation, neither of which is particularl difficult to overcome if ou take time to think about the problem before ou start taking the derivative.. Be careful to appl the product and quotient rules whenever appropriate. Ever time ou have a mixture of x's an's in a single term ou will have to use one of the above rules. To take the derivative of, ou need to use the product rule on the left side: x 4 d d 0 x* * x factor, then ou will have to have a / next to it. In this example we would have When ou actuall take the derivatives, if the factor is a - x* *x 0.. Ever time ou take the derivative of a, ou must write down a /. You have alrea been doing this, although ou ma have used the notation ' instead anou also ma have not even noticed what ou were doing. For example: If = x - x + 5 anou take the derivative, ou will get ' = / = x + (remember that the notations ' and / are interchangeable). You have actuall taken the derivative of the on the left side of the original equation, and indicated that b either writing * ' or * / (the notation which sas "the derivative of with respect to x"). You have done this unconsciousl up until now. Now ou will have to remember to do that same thing ever time ou come across a and take its derivative because the 's will not be b themselves on one side of the equation.

2 Taking the derivative with implicit differentiation is reall a form of using the chain rule. A good rule of thumb is to remember that if ou are taking the derivative in terms where the variables d x x agree then ou have no problem:. d If the variables do not agree, then use the chain rule, as in. If ou have a mixture of variables, then ou have to use the product or quotient rule: d d d x x x x x Our process has four steps: Implicit Differentiation Process. Differentiate both sides of the equation with respect to x.. Collect all the / terms on one side of the equation and move the rest of the terms over to the other side.. Factor out the /. 4. Solve for / b dividing. Important: If ou don't have an / terms when ou finish step # ou have done something wrong!!!!

3 Implicit Differentiation Continued Let s look at some examples. Remember that our process has four steps: Implicit Differentiation Process. Differetiate both sides of the equation with respect to x.. Collect all the / terms on one side of the equation and move the rest of the terms over to the other side.. Factor out the /. 4. Solve for / b dividing. Important: If ou don't have an / terms or an = sign when ou finish step # ou have done something wrong!!!! xx 6 A. We cannot easil solve for in this equation, so we will use implicit differentiation. xx 6 x *x x * 0 0 x x x x x x x x x x x x First, let's look at it term b term to see what we have to do to find the derivative: x : this is the product of x an, so we use the product rule: (st) * (deriv of nd) + (nd) * (deriv of st) x : this is the product of x an, so we use the product rule again 6 is a constant so its derivative will be zero Now we go from left to right and take the derivative: [derivative of x ] + [derivative of x ] = derivative of 6 Notice that ever time I took the derivative of a, I wrote down a /. Now we just simplif a bit. Put the terms with / on the same side and move the other terms over. Factor the / out. Divide to solve for /. You can see right awa that ou will need to keep ourself organized when ou do these, or ou will forget a part of the problem!

4 #B. x + sin = x Again, there is no wa to solve for, so we have to use implicit differentiation. x + sin = x cos x* * cos x cos x cos x Look through the problem to see what we will have to do: x: This is an x-term, so I take the derivative normall. sin : This is a -term, so when I take the derivative I must have a /. x: This is the product rule with x *. When I take the derivative of the, I must have a /. Take the derivative. Check ourself here. Do still have our = sign in the same place? Do ou have a / in the terms where ou took the derivative of the? Now I have to gather the / terms on one side. Factor out the /. Divide to solve for /. Some of ou might be able to do the last two steps in our head, and that is fine if ou are rea to take that shortcut. Just make sure ou are correctl solving for /.

5 #C: Given the function x 6, find the value of at the point (,). We will be taking the derivative twice and then evaluating it at the given point. Even though I could solve for here, it will be much easier to use implicit differentiation. Resist the urge to solve for!! x x 6 Look through the problem to see what we will have to do: 0 x x x x x x 4 x : This is an x-term, so I take the derivative normall. : This is a -term, so when I take the derivative I must have a /. 6: This a constant with derivative of 0. Take the derivative. Check ourself here. Do still have our = sign in the same place? Do ou have a / in the term where ou took the derivative of the? Now solve for /. To take the second derivative I am going to use the quotient rule: top = x and bottom =. Remember to write a / when ou take the derivative of the. Now, what a mess this seems to be! But, since we are asked to evaluate this at the point (,), all we reall have to do from here is sub in the values that we know. Put in x =, =, and / = - (evaluate the / we got in a previous step), and solve for the second derivative. If I had not been evaluating the second derivative, I would have had to simplif some b putting the value for / into the derivative and then simplifing. I will do it here so that ou can see the process. x x x 4 x The moral of this is to alwas stop and think about what ou need to do. If ou are not told to solve for, then don t; just evaluate instead. No sense in doing extra work! You don t have to simplif more than this. 4 x x 4

6 #D: We have to find the equations of the tangent and the normal lines to the curve x 9 at the point (-, ). Since to find the equation of a line we must have a slope we will take the derivative to find the slope of the tangent line. The normal line is perpendicular to the tangent, so its slope will be the negative reciprocal of that of the tangent line. x 9 Looking at the problem we see that we have to use the product x x x at (-,) m t = tangent line: - = (x + ) normal line: at (-,) m n = -/ x x 0 - = -/ (x + ) rule, with u = x and v =. Take the derivative with the product rule on the left side and remember the / when ou take the derivative of the! Now solve for /, which is the formula for our slope. We can now find the slopes of our lines and find their equations. For the tangent line: Evaluate the slope at the given point first. Write the equation. For the normal line: Take the negative reciprocal of the tangent's slope. Write the equation. Note on this problem - here, it would have been OK to rewrite the problem first: 9x and then take the derivative. It would have worked the same wa. The first derivative would have been 8x. Then ou could sub in the coordinates of the point to get m => () / = -8 (-) - => m = for the tangent line, etc. Remember, that implicit differentiation is not reall an harder than an other methods, but ou will have to work carefull. And don't forget the / and don t pick up our = sign and move it around!!! Don't be one of those students who gets through the first step of taking the derivative and never notices that there are no / terms or that the = sign is moved! If ou don't have an /, then something went wrong anou need to go back and check to be sure that ever time ou took the derivative of a -term, ou wrote down a /. Do a quick check for the = sign. If it is not in the same place, then ou need to go back and fix it. 4

7 Homework Examples #5, 9, #5: Use implicit differentiation to find / and for x x x x This is our function. The do not want us to solve for (it would make the problem harder anwa!). x Take the derivative implicitl. x x () ( x) ( x ) x Solve for /. I simplified a bit. Take the derivative again with the quotient rule. Sub in the value for / from the first step and simplif. ( x ) ( ) x #9: Find the lines that are tangent and normal to the curve x 9 at (-, ). x 9 Here is the function. ) 0 x x x x (x At (-, ) m = tangent line: - = (x + ) For the normal line: m = -/ - = (-/) (x + ) Take the derivative implicitl. Be careful! On the left side we have x * so we have to use the product rule. Solve for /. Evaluate the derivative to find the slope of the tangent line. Write the equation. The normal line is perpendicular to the tangent.

8 #: Find the lines that are tangent and normal to the curve x sin at,. x sin Here is the function. x () cos 0 x cos 0 Take the derivative implicitl. Don't forget to use the product rule on the first term (we have x* ). Also, remember that is a constant. x cos At, * m () cos tangent line: ( x ) For the normal line: m ( x ) Solve for /. Evaluate the derivative to find the slope of the tangent line. Write the equation. Don t forget to label the equation as the tangent line! The reader will not guess which one ou wanted it to be!! The normal line is perpendicular to the tangent.

9 Homework Examples #5, 9, 7, #5: Find / for the function / x5 x 5 / 5 x / ( ) This is our function. Take the derivative with the general power rule. Be careful to subtract correctl when ou find the new exponent! x 5 / #9: Find / for the function xx 6 x x 6 This is our function. x (x) x () 0 0 x x x x x x x x x x x x Take the derivative. We have to use implicit differentiation this time. The first term requires the product rule: x * The second term requires the product rule: x * The 6 is a constant. Simplif a bit and then gather the / terms together. Factor out the / and then divide to solve for /. #7: Find / for the function x = tan x tan This is our function sec cos Take the derivative. We have to use implicit differentiation again. Isolate the / term on one side. I divided both sides b and then used the trig identit cos sec sec

10 #: which of the following could be true if a) c) / f( x) x b) 4/ f ( x) x d) / ( )? f x x 9 f x 0 / f( x) x 6 5/ ( ) x 7 We will have to check each of the answers because there ma be more than one answer. / f( x) x f ( x) x x a) Not true b) f x 9 0 5/ ( ) x 7 / / 9 5 f ( x) x x 0 / / f ( x) x x True / / If this is true, we should be able to take the derivative twice and get our given function. We have the right exponent after onl one derivative, so we know it is not true. If this is true, we should be able to take the derivative twice and get our given function. This one works. c) f ( x) x 4/ If this is true, we should be able to start with given function and take its derivative to get (c). f ( x) x f ( x) x If True d) / 4/ f x x / ( ) 6 f ( x) x x True / / This works too. If this is true, we should be able to take the derivative and get our given function. This one works too.