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3.7 – Implicit Differentiation

3.7 – Implicit Differentiation. Consider the equations x y = 1 x 2 + y 2 = 4 x 3 + y 3 – 3xy = 0 What do they all have in common? They are all defined implicitly , or we write them with x’s and y’s on the same side.

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3.7 – Implicit Differentiation

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  1. 3.7 – Implicit Differentiation Consider the equations xy = 1 x2 + y2 = 4 x3 + y3 – 3xy = 0 What do they all have in common? They are all defined implicitly, or we write them with x’s and y’s on the same side.

  2. The first is a hyperbola, which we can write explicitly as y = 1/x. The second a circle centered at the origin with radius 4 we can re-write as Notice the ±. This means if we are to graph it (in our calculators) we must enter two separate functions. The + square root, and the -. Notice, now that we’ve solved for y, we can find a derivative at any point.

  3. x3 + y3 – 3xy = 0 The third equation, called a Folium of Descartes, is not nearly as easy to solve for y. However, if you could, you take a derivative like the previous examples, using the appropriate piece, no problem. But remember, we can’t.

  4. This could be a problem. This is, if we weren’t intelligent, hard-working calculus students. When we encounter equations like this, we need to use implicit differentiation.

  5. Example 1 Simple example. Find dy/dx for y2 = x First, take the derivative of both sides, with respect to x. 2y = 1. (this seems like it would be ok, but its not.) Whenever we differentiate a non x term (w/ respect to x) we need to multiply by a factor of dy/dx. So the above should look like 2y dy/dx = 1 The last step is to solve for dy/dx by dividing by 2y. dy/dx = 1/(2y) Done.

  6. Example 2 Going back to the circle from the beginning, x2 + y2 = 4 Differentiate both sides w/r to x. 2x + 2y dy/dx = 0 (keep all terms w/ dy/dx on one side) 2y dy/dx = -2x (divide) dy/dx = -2x/2y = -x/y

  7. Example 2a Finding a slope – Find the slope of the line tangent to the previous circle at (2√2,2√2) • We know dy/dx = -x/y • Plug in x and y (-2√2)/(2√2) = -1 Find the slope of the line at (1, -√3) -1 / -√3 = √3/3

  8. Example 3 Finding a 2nd derivative implicitly. Find if 2x3 – 3y2 = 8 First, find the first derivative ! (dy/dx) 6x2 – 6y dy/dx = 0 -6y dy/dx = -6x2 dy/dx = -6x2/-6y = x2/y (of course, y≠0) We’re half way there, wooooo-oooooo (loap)

  9. We already found dy/dx = x2/y To find d2y/dx2, we’ll need the quotient rule. Refresh (u’v – uv’)/v2. Ready? d2y/dx2= Next, replace dy/dx with x2/y =

  10. a little bit of algebra later = Finally,

  11. Be able to use the product and quotient rules, chain rule with implicit differentiation. Do the HW!

  12. New (Old) Topic Using the power rule with rational exponents. Recall Examples. Find the derivative of each • y = x2/3 • y = 1/x4/5

  13. Solutions a) y’ = b) Re-write as y = x-4/5

  14. Homework

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