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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