Chapter 3 | Derivatives
313
⎛ ⎝ − x y
⎞ ⎠
d 2 y dx 2
dy dx = −
= d
x y .
Differentiate both sides of
dy
⎛ ⎝ 1· y − x dy dx
⎞ ⎠
Use the quotient rule to find d dy ⎛ ⎝ − x y ⎞ ⎠ .
= −
y 2
− y + x dy dx y 2 − y + x ⎛
=
Simplify.
⎞ ⎠
⎝ − x y
dy dx = −
x y .
=
Substitute
y 2
y 2 − x 2 y 3
= −
Simplify.
d 2 y dx 2 . If we choose, we can simplify the expression further by
At this point we have found an expression for
recalling that x 2 + y 2 =25 and making this substitution in the numerator to obtain d 2 y dx 2
= − 25 y 3 .
dy dx for y defined implicitly by the equation 4 x 5 +tan y = y 2 +5 x .
3.48
Find
Finding Tangent Lines Implicitly Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations. Example 3.71 Finding a Tangent Line to a Circle
Find the equation of the line tangent to the curve x 2 + y 2 =25 at the point (3, −4).
Solution Although we could find this equation without using implicit differentiation, using that method makes it much easier. In Example 3.68 , we found dy dx = − x y . The slope of the tangent line is found by substituting (3, −4) into this expression. Consequently, the slope of the tangent line is dy dx | (3, −4) = − 3 −4 = 3 4 . Using the point (3, −4) and the slope 3 4 in the point-slope equation of the line, we obtain the equation y = 3 4 x − 25 4 ( Figure 3.31 ).
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