SOLUTIONS TO LOGARITHMIC DIFFERENTIATION


SOLUTION 1 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

y = xx .

Apply the natural logarithm to both sides of this equation getting

$ \ln y = \ln x^x $

$ = x \ln x $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule on the right-hand side. Thus, beginning with

$ \ln y = x \ln x $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = x \displaystyle{ 1 \over x } + (1) \ln x $

$ = 1 + \ln x $ .

Multiply both sides of this equation by y, getting

$ y' = y (1 + \ln x) = x^x (1 + \ln x) $ .

Click HERE to return to the list of problems.



SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

y = x(ex) .

Apply the natural logarithm to both sides of this equation getting

$ \ln y = \ln x^{(e^x)} $

$ = e^x \ln x $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule on the right-hand side. Thus, beginning with

$ \ln y = e^x \ln x $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = e^x \Big\{ \displaystyle{ 1 \over x } \Big\} + e^x \ln x $

(Get a common denominator and combine fractions on the right-hand side.)

$ = \displaystyle{ e^x \over x } + e^x \ln x \Big\{ \displaystyle{ x \over x } \Big\} $

$ = \displaystyle{ e^x \over x } + \displaystyle{ x e^x \ln x \over x } $

$ = \displaystyle{e^x + x e^x \ln x \over x } $

(Factor out ex in the numerator.)

$ = \displaystyle{e^x (1 + x \ln x ) \over x } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{e^x (1 + x \ln x ) \over x } $

$ = x^{(e^x)} \displaystyle{e^x (1 + x \ln x ) \over x^1 } $

(Combine the powers of x .)

$ = x^{(e^x-1)} e^x (1 + x \ln x ) $ .

Click HERE to return to the list of problems.



SOLUTION 3 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

y = (3x2+5)1/x .

Apply the natural logarithm to both sides of this equation getting

$ \ln y = \ln (3x^2+5)^{1/x} $

$ = (1/x) \ln (3x^2+5) $

$ = \displaystyle{ \ln (3x^2+5) \over x } $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the quotient rule and the chain rule on the right-hand side. Thus, beginning with

$ \ln y = \displaystyle{ \ln (3x^2+5) \over x } $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' =
\displaystyle{ x \Big\{ \displaystyle{ 1 \over 3x^2+5 } \Big\} (6x) - \ln(3x^2+5) (1) \over x^2 } $

(Get a common denominator and combine fractions in the numerator.)

$ = \displaystyle{ \displaystyle{ 6x^2 \over 3x^2+5 } -
\ln(3x^2+5) \Big\{ \displaystyle{ 3x^2+5 \over 3x^2+5 } \Big\} \over \displaystyle{ x^2 \over 1 } } $

(Dividing by a fraction is the same as multiplying by its reciprocal.)

$ = \displaystyle{ 6x^2 - (3x^2+5) \ln(3x^2+5) \over 3x^2+5 } \displaystyle{ 1 \over x^2 } $

$ = \displaystyle{ 6x^2 - (3x^2+5) \ln(3x^2+5) \over x^2 (3x^2+5) } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{ 6x^2 - (3x^2+5) \ln(3x^2+5) \over x^2 (3x^2+5) } $

$ = (3x^2+5)^{1/x} \displaystyle{ 6x^2 - (3x^2+5) \ln(3x^2+5) \over x^2 (3x^2+5)^1 } $

(Combine the powers of (3x2+5) .)

$ = \displaystyle{ (3x^2+5)^{(1/x-1)} \big\{ 6x^2 - (3x^2+5) \ln(3x^2+5) \big\} \over x^2 } $ .

Click HERE to return to the list of problems.



SOLUTION 4 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

$ y = (\sin x)^{x^3} $ .

Apply the natural logarithm to both sides of this equation getting

$ \ln y = \ln (\sin x)^{x^3} $

$ = x^3 \ln (\sin x) $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule and the chain rule on the right-hand side. Thus, beginning with truein $ \ln y = x^3 \ln (\sin x) $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = x^3 \Big\{ \displaystyle{ 1 \over \sin x } \Big\} \cos x + (3x^2) \ln (\sin x) $

(Get a common denominator and combine fractions on the right-hand side.)

$ \displaystyle{ { 1 \over y } } y' =
\displaystyle{ x^3\cos x \over \sin x } + 3x^2 \ln (\sin x) \Big\{ \displaystyle{ \sin x \over \sin x } \Big\} $

$ = \displaystyle{ x^3 \cos x + 3x^2 \sin x \ln(\sin x) \over \sin x } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{ x^3\cos x + 3x^2 \sin x \ln (\sin x) \over \sin x } $

$ = (\sin x)^{x^3} \displaystyle{ x^3\cos x + 3x^2 \sin x \ln (\sin x) \over (\sin x)^1 } $

(Combine the powers of $ (\sin x) $ .)

$ = (\sin x)^{(x^3-1)} \big\{ x^3\cos x + 3x^2 \sin x \ln (\sin x) \big\} $ .

Click HERE to return to the list of problems.



SOLUTION 5 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

$ y = 7x (\cos x)^{x/2} $ .

Apply the natural logarithm to both sides of this equation and use the algebraic properties of logarithms, getting

$ \ln y = \ln \Big( (7x) (\cos x)^{x/2} \Big) $

$ = \ln (7x) + \ln (\cos x)^{x/2} $

$ = \ln (7x) + (x/2) \ln (\cos x) $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule and the chain rule on the right-hand side. Thus, beginning with

$ \ln y = \ln (7x) + (x/2) \ln (\cos x) $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = \Big\{ \displaystyle{ 1 \over 7x } \Big\} ...
...) \Big\{ \displaystyle{ 1 \over \cos x } \Big\} (-\sin x) + (1/2) \ln (\cos x) $

$ = \displaystyle{ 1 \over x }
- \displaystyle{ x \sin x \over 2 \cos x } + \displaystyle{ \ln (\cos x) \over 2 } $

(Get a common denominator and combine fractions on the right-hand side.)

$ = \displaystyle{ 1 \over x } \Big\{ \displaystyle{ 2 \cos x \over 2 \cos x } \...
... \ln (\cos x) \over 2 } \Big\{ \displaystyle{ x \cos x \over x \cos x } \Big\} $

$ = \displaystyle{ 2 \cos x - x^2 \sin x + x \cos x \ln (\cos x) \over 2x \cos x } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{ 2 \cos x - x^2 \sin x + x \cos x \ln (\cos x) \over 2x \cos x } $

$ = 7x (\cos x)^{x/2} \displaystyle{ 2 \cos x - x^2 \sin x + x \cos x \ln (\cos x) \over 2x \cos x } $

(Divide out a factor of x .)

$ = 7 (\cos x)^{x/2} \displaystyle{ 2 \cos x - x^2 \sin x + x \cos x \ln (\cos x) \over 2 (\cos x)^1 } $

(Combine the powers of $ (\cos x) $ .)

$ = (7/2) (\cos x)^{(x/2-1)} \big\{ 2 \cos x - x^2 \sin x + x \cos x \ln (\cos x) \big\} $ .

Click HERE to return to the list of problems.



SOLUTION 6 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

$ y = \sqrt{x}^{ \sqrt{x} } e^{ x^2 } $ .

Apply the natural logarithm to both sides of this equation and use the algebraic properties of logarithms, getting

$ \ln y = \ln \Big( \sqrt{x}^{ \sqrt{x} } e^{ x^2 } \Big) $

$ = \ln \Big( \sqrt{x}^{ \sqrt{x} } \Big) + \ln \Big( e^{ x^2 } \Big) $

$ = \sqrt{x} \ln ( \sqrt{x} ) + x^2 \ln ( e ) $

$ = \sqrt{x} \ln ( \sqrt{x} ) + x^2 ( 1 ) $

$ = \sqrt{x} \ln ( \sqrt{x} ) + x^2 $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule and the chain rule on the right-hand side. Thus, beginning with

$ \ln y = \sqrt{x} \ln ( \sqrt{x} ) + x^2 $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = \sqrt{x} \Big\{ \displaystyle{ 1 \over \sqrt{x} } \Big\} (1/2)x^{-1/2}
+ (1/2)x^{-1/2} \ln ( \sqrt{x} ) + 2x $

$ = \displaystyle{ 1 \over 2 \sqrt{x} } + \displaystyle{ \ln ( \sqrt{x} ) \over 2 \sqrt{x} } + 2x $

(Get a common denominator and combine fractions on the right-hand side.)

$ = \displaystyle{ 1 \over 2 \sqrt{x} } + \displaystyle{ \ln ( \sqrt{x} ) \over 2 \sqrt{x} }
+ 2x \Big\{ \displaystyle{ 2 \sqrt{x} \over 2 \sqrt{x} } \Big\} $

$ = \displaystyle{ 1 + \ln ( \sqrt{x} ) + 4 x^{ 1+1/2 } \over 2 \sqrt{x} } $

$ = \displaystyle{ 1 + \ln ( \sqrt{x} ) + 4 x^{ 3/2 } \over 2 \sqrt{x} } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{ 1 + \ln ( \sqrt{x} ) + 4 x^{ 3/2 } \over 2 \sqrt{x} } $

$ = \sqrt{x}^{ \sqrt{x} } e^{ x^2 } \displaystyle{ 1 + \ln ( \sqrt{x} ) + 4 x^{ 3/2 } \over 2 \sqrt{x}^1 } $

(Combine the powers of $ \sqrt{x} $ .)

$ = (1/2) \sqrt{x}^{( \sqrt{x} - 1) } e^{ x^2 } \big\{ 1 + \ln ( \sqrt{x} ) + 4 x^{ 3/2 } \big\} $ .

Click HERE to return to the list of problems.



SOLUTION 7 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! The function must first be revised before a derivative can be taken. Begin with

$ y = x^{ \ln x } (\sec x)^{3x} $ .

Apply the natural logarithm to both sides of this equation and use the algebraic properties of logarithms, getting

$ \ln y = \ln \Big( x^{ \ln x } (\sec x)^{3x} \Big) $

$ = \ln x^{ (\ln x) } + \ln (\sec x)^{3x} $

$ = (\ln x) (\ln x) + 3x \ln ( \sec x ) $

$ = ( \ln x )^2 + 3x \ln ( \sec x ) $ .

Differentiate both sides of this equation. The left-hand side requires the chain rule since y represents a function of x . Use the product rule and the chain rule on the right-hand side. Thus, beginning with

$ \ln y = ( \ln x )^2 + (3x) \ln ( \sec x ) $

and differentiating, we get

$ \displaystyle{ { 1 \over y } } y' = 2 ( \ln x ) \Big\{ \displaystyle{ 1 \over ...
... \displaystyle{ 1 \over \sec x } \Big\} ( \sec x \tan x ) + (3) \ln ( \sec x ) $

(Divide out a factor of $ \sec x $ .)

$ = \displaystyle{ 2 \ln x \over x } + 3x \tan x + 3 \ln ( \sec x ) $

(Get a common denominator and combine fractions on the right-hand side.)

$ = \displaystyle{ 2 \ln x \over x } + 3x \tan x \Big\{ \displaystyle{ x \over x } \Big\}
+ 3 \ln ( \sec x ) \Big\{ \displaystyle{ x \over x } \Big\} $

$ = \displaystyle{ 2 \ln x + 3x^2 \tan x + 3x \ln ( \sec x ) \over x } $ .

Multiply both sides of this equation by y, getting

$ y' = y \displaystyle{ 2 \ln x + 3x^2 \tan x + 3x \ln ( \sec x ) \over x } $

$ = x^{ \ln x } (\sec x)^{3x} \displaystyle{ 2 \ln x + 3x^2 \tan x + 3x \ln ( \sec x ) \over x^1 } $

(Combine the powers of x .)

$ = x^{ (\ln x - 1) } (\sec x)^{3x} \big\{ 2 \ln x + 3x^2 \tan x + 3x \ln ( \sec x ) \big\} $

Click HERE to return to the list of problems.


.



 

Duane Kouba
1998-06-06