IMPLICIT DIFFERENTIATION EXAMPLE PROBLEMS

A function in which the dependent variable is expressed solely in terms of the independent variable x, namely, y = f(x), is said to be an explicit function. For instance, y = (1/2)x3 - 1 is an explicit function, whereas an equivalent equation 2y − x3 + 2 = 0 is said to define the function implicitly or y is an implicit function of x.

Question 1 :

Find the derivatives of the following

√(x2 + y2)  =  tan-1(y/x)

Solution :

√(x2 + y2)  =  tan-1(y/x)

By finding derivation on left hand side, we get

  =  (1/2√(x2+y2))(2x+2y(dy/dx))

  =  [x+y(dy/dx)]/√(x2+y2)  ----(1)

By finding derivation on right left hand side, we get

  =  1/[1 + (y/x)2]  (-y/x2) + (1/x) (dy/dx) 

  =  x2/(x2 + y2(x (dy/dx) - y)/x2

  =  (x (dy/dx) - y)/(x2 + y2)  ----(2)

(1)  =  (2)

By equating (1) and (2), we get

[x+y(dy/dx)]/√(x2+y2)  =  (x (dy/dx) - y)/(x2 + y2)

√(x2+y2)[x+y(dy/dx)]  =  (x (dy/dx) - y)

x√(x2+y2) + y √(x2+y2)(dy/dx) - x (dy/dx)  =  -y

(dy/dx) (y√(x2+y2) - x)  =  -y-x√(x2+y2)

(dy/dx)  =  (x√(x2+y2) + y) / (x - y√(x2+y2))

Question 2 :

Differentiate the following

tan (x + y) + tan (x - y) = x

Solution :

tan (x + y) + tan (x - y) = x

Differentiate with respect to "x"

sec2(x + y) (1 + dy/dx) + sec2(x - y) (1 - dy/dx)  =  1

sec2(x+y)+sec2(x+y)(dy/dx)=1-sec2(x-y)-sec2(x-y)(dy/dx)

sec2(x+y)(dy/dx)+sec2(x-y)(dy/dx) = 1-sec2(x-y)-sec2(x-y)

(dy/dx) (sec2(x+y)+sec2(x-y))  =  1 - sec2(x-y)- sec2(x-y)

dy/dx  =  (1 - sec2(x-y)- sec2(x-y)) / (sec2(x+y)+sec2(x-y))

Question 3 :

Differentiate the following

If cos (xy) = x, show that dy/dx  =  -(1+ysin(xy))/x sin(xy)

Solution :

cos (xy) = x

Differentiate with respect to "x"

-sin (xy) [x(dy/dx) + y(1)]  =  1

-sin (xy)(dy/dx) - y sin (xy)  =  1

-sin (xy)(dy/dx) = 1 + ysin(xy)

dy/dx  =  -[1 + ysin(xy)]/sin (xy)

Question 4 :

Differentiate the following

tan-1[√(1 - cos x)/(1+cosx)]

Solution :

Let y  =  tan-1[√(1 - cos x)/(1+cosx)]

Trigonometric formula for (1-cos x)/(1+cosx)  =  tan2(x/2)

By applying trigonometric formula, we may find derivatives easily. 

 y  =  tan-1[tan2(x/2)]

y = x/2

Differentiating with respect to "x"

dy/dx  =  1/2

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