Karl's Calculus Tutor - Solution to Exercise 7.2-4

Solution to Exercise 7.2-4KCT logo

© 1998 by Karl Hahn

I'm not going to go into a lot of explanation on these. If you need to review implicit differentiation, the product rule, or the chain rule, you should do so, and be serious about it. Those concepts will be on the exam. 

a)   sin(y) = cos(x2)
Use implicit differentiation on the left and the chain rule on the right. 
   cos(y) y'  =  -sin(x2) (2x)


b)   y tan(x) = x
Use implicit differentiation combined with the product rule on the left. The right is trivial. 
   y' tan(x)  +  y (1 + tan2(x))  =  1
or equivalently 
   y' tan(x)  +  y sec2(x)  =  1


c)   sin(xy)cos(x) = y2
You have to use the product rule twice, the chain rule, and implicit differentiation on the left. The right just requires implicit differentiation. Using implicit differentiation and the product rule on xy gives its derivative to be xy' + y. That is what the chain rule will bring out when we take the derivative of sin(xy). Hence its derivative is cos(xy) (xy' + y). From there you apply the product rule to the entire left side. The final result is 
   -sin(xy)sin(x)  +  cos(xy) (xy' + y) cos(x)  =  2y y'


d)   tan2(y) = cos(2x)
Apply the chain rule and implicit differentiation to the left and the chain rule to the right. 
   2tan(y) (1 + tan2(y)) y'  =  -2 sin(2x)

or equivalently 
   2tan(y) sec2(y) y'  =  -2 sin(2x)

Return to Main Text

email me at hahn@netsrq.com