# We differentiate both sides of the given equation

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We differentiate both sides of the given equation with respect to x . The most complicated part of this is that we have to use the chain rule in order to differentiate 2 y with respect to x . In particular, we obtain:
2 2 2 2 25 25 2 2 0 2 2 d d dy x y x y x y dx dx dx dy dy x y x dx dx y  as long as 0 y ( Note: From the diagram in the motivation we see we have a vertical tangent when y is zero) . (b) Find an equation of the tangent line to the circle given by 2 2 25 x y at the point 3,4 . We know that the slope of the desired tangent line can by found from the first derivative. Specifically, the slope of our tangent line will be: 3 4 3 4 x y dy dx . Thus, an acceptable answer is: 3 3 25 4 3 4 4 4 y x y x . Note: We could have done part (b) in the above example by first solving for y (and only keep the function for y that leads to the point (3,4)). Specifically, 2 25 y x . Then we could have used this formula to differentiate and complete part (b). However, notice that our solution via implicit differentiation is less messy. In fact when solving for y is impossible, implicit differentiation is our only option! Graded Example: Find the first derivative of y with respect to x if 2 sin cos x y y x . Solution: Following the idea of implicit differentiation, we have: 2 2 2 2 2 2 2 2 sin cos sin cos cos cos cos cos 1 sin 2 cos cos 2 cos sin cos cos 2 cos sin cos sin cos cos d d x y y x x y y x dx dx d d d x y x y y x y x dx dx dx dy dy x y y x y x dx dx dy dy x y y x y x x y dx dx dy x y y x y x x y dx y x x y dy dx x    2 sin cos . 2 cos 2 cos cos y x x y y y x y x x y
Recommended Homework: 3.4: 25-47 odd; 55; 69; 77; 87; 97 3.5: 5-15 odd
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