LOGARITHMIC DIFFERENTIATION PROBLEMS AND SOLUTIONS

Differentiate each of the following with respect to x.

Problem 1 :

x^y = y^x 

Solution :

x^y = y^x 

Taking logarithm on both sides.

logx^y = logy^x

ylogx = xlogy

Differentiate with respect to x. 

y(1/x) + logx(dy/dx) = x(1/y)(dy/dx) + logy(1)

y/x + logx(dy/dx) = (x/y)(dy/dx) + logy

logx(dy/dx) - (x/y)(dy/dx) = logy - y/x

(logx - x/y)(dy/dx) = (xlogy - y)/x

[(ylogx - x)/y](dy/dx) = (xlogy - y)/x

dy/dx = (y/x)[(xlogy - y)/(ylogx - x)]

Problem 2 :

y = (cosx)^logx

Solution :

y = (cosx)^logx

logy = log[(cosx)^logx]

logy = (logx)log(cosx)

Differentiate with respect to x. 

(1/y)(dy/dx) = logx(1/cosx)(-sinx) + log(cosx)(1/x)

(1/y)(dy/dx) = -logxtanx + log(cosx)/x

(dy/dx) = y[-logxtanx + log(cosx)/x]

dy/dx = (cosx)^logx[-logxtanx + log(cos x)/x]

Problem 3 :

(x²/a²) + (y²/b²) = 1

Solution :

Its an implicit function. Since the function is not a complicated one, we don't have to use logarithm to find derivative.

(x²/a²) + (y²/b²) = 1

Differentiate with respect to x. 

(2x/a²) + (2y/b²)(dy/dx) = 0

(2y/b²)(dy/dx) = -(2x/a²)

(dy/dx) = -(b²/a²)(x/y)

(dy/dx) = -(b²x/a²y)

Problem 4 :

√(x² + y²) = tan^-1(y/x)

Solution :

√(x² + y²) = tan^-1(y/x)

Differentiate √(x² + y²) with respect to x.

= [1/2√(x²+ y²)][2x + 2y(dy/dx)]

= [x + y(dy/dx)]/√(x²+y²) ----(1)

Differentiate tan^-1(y/x) with respect to x.

  = 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)

[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))

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