
To calculate: The derivative of the function tan(x−y)=y1+x2 by implicit

Answer to Problem 14E
The derivative of the function is 2xy+(1+x2)2sec2(x−y)1+x2+(1+x2)2sec2(x−y) .
Explanation of Solution
Given information:
The function tan(x−y)=y1+x2 .
Formula used:
Thechain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .
Power rule for differentiation is ddxxn=nxn−1 .
Product rule for differentiation is ddx(f⋅g)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .
Derivative of tanx is sec2x .
Calculation:
Consider the function tan(x−y)=y1+x2 .
Differentiate both sides with respect to x ,
ddx(tan(x−y))=ddx(y1+x2)
Recall that power rule for differentiation is ddxxn=nxn−1 and chain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .
Also for the terms of the above expression, apply the product rule for differentiation.
Recall that quotient rule for differentiation is ddx(fg)=g(x)f'(x)−f(x)g'(x)[g(x)]2 where f and g are functions of x .
Apply it. Also observe that y is a function of x,
ddx(tan(x−y))=ddx(y1+x2)sec2(x−y)[1−y']=(1+x2)y'−y(2x)(1+x2)2(1+x2)2sec2(x−y)−(1+x2)2sec2(x−y)y'=y'+x2y'−2xy
Isolate the value of y' on left hand side and simplify,
(1+x2)2sec2(x−y)−(1+x2)2sec2(x−y)y'=y'+x2y'−2xy(−x2−1−(1+x2)2sec2(x−y))y'=−2xy−(1+x2)2sec2(x−y)y'=−2xy−(1+x2)2sec2(x−y)−1−x2−(1+x2)2sec2(x−y)y'=2xy+(1+x2)2sec2(x−y)1+x2+(1+x2)2sec2(x−y)
Thus, the derivative of the function is 2xy+(1+x2)2sec2(x−y)1+x2+(1+x2)2sec2(x−y) .
Chapter 3 Solutions
Single Variable Calculus: Concepts and Contexts, Enhanced Edition
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