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 = x^x .

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

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

$= 1 + \ln x$ .

Multiply both sides of this equation by y, getting

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

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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

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

(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 e^x 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 .)

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

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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

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

(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) .)

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

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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

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

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

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

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

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 $(\sin x)$ .)

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

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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

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

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

(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 $(\cos x)$ .)

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

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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

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

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

(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 $\sqrt{x}$ .)

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

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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

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

(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 .)

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

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