Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

Chain Rule

  • When finding the derivative of a composite function, take the derivative of the outside function with the inside function g considered as the variable, leaving the “inside” function alone. Then we multiply this by the derivative of the inside function, with respect to the variable x.

    • The Chain Rule: If y = f(g(x)), then y’ = f’(g(x)) * g’(x) OR
    • If y = y(v) and v = v(x), then dy/dx = (dy/dv)*(dv/dx)
  • My personal memory trick for this is “douter, inner, dinner) → drop the power down to outside the parathesis, leave the inner, multiply by the derivative of the inner

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Implicit Differentiation

  • When you can’t isolate y in terms of x, you take the derivative implicitly. Essentially, you solve for the derivative of x with respect to y, in order to get a derivative in terms of both variables.

    • (dx/dy) = (1/(dy/dx))
  • Solving with the reciprocal allows to split up the variable and pair it to both sides, so that they can be factored.

    • You can also apply other, previous rules like the product rule in order to solve it. The AP exam loves to do this.
  • An easier way of describing implicit differentiation is that if your variable doesn’t match dx, then you need to follow it up with d(variable)/dx

  • For example, if we’re given x^2 + y^2 = 25 at the point (3, 4), we need to implicitly differentiate → doing this with respect to x we get:

    • 2x + 2y(dy/dx) = 0
  • Then you have to solve for dy/dx

    • dy/dx = -x/y
  • At the point (3, 4), we have x = 3 and y = 4. Substituting these values, we get:

  • dy/dx = -3/4

    • So the slope of the tangent line is -3/4.
    • We also know that the point (3, 4) lies on the tangent line. Using the point-slope form of the equation of a line, we get:
  • y - 4 = -3/4(x - 3)

    • This is the equation of our tangent line

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Inverse Function Differentiation

  • There is a simple formula in order to find the derivative of an inverse function.

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  • In short, we can find the derivative at a particular point by taking the reciprocal of the derivative at that point’s corresponding y value.

    • Say you want the derivative at the point (1,2) - you want g’(1). So find f’(2), then take the reciprocal - this is the value of g’(1).
    • Our original point is (1,2) so our reciprocal will be (2,1)
    • The same goes for our slope, find the slope at f’(2) and then take the reciprocal
    • Remember, f’(x) is the same as the slope!
  • The AP test usually only has 1-2 of these questions so don’t stress too much! đź‘Ť

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Inverse Trigonometry

  • This is going to be one that is easier to just memorize, but you can also find them by following the formulas explained in implicit differentiation and using trigonometry rules.

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HINTS

  • When two terms are multiplied together, use product rule unless it’s easier to multiply it out
  • If you see a function within another function, you will almost certainly have to use chain rule
  • If there are x and y terms mixed together, we will need to use implicit differentiation
    • If you’re finding the derivative at a point, just plug it in and avoid the solving out
  • When evaluating derivatives at a point, look to see if the terms become one or zero
  • You can mentally take certain derivatives
  • If it is required to take a second derivative, simplify the first derivative before you start

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