Exponential, Logarithmic Functions and Logarithmic Differentiation

Exponential and Logarithmic Functions

The Exponential Function

For a positive base $b>0,\ b\neq 1$, the exponential function is $$y=b^x.$$ In calculus, the most important base is the mathematical constant $e\approx 2.718$. The function $$y=e^x$$ is called the natural exponential function.

For $y=e^x$:

• Domain: $\mathbb{R}$
• Range: $(0,\infty)$

Key properties

• It always passes through $(0,1)$ because $e^0=1$.
• It is always positive.
• It is strictly increasing.
• Its derivative is the function itself.

The Logarithmic Function

If $b>0,\ b\neq 1$, then $$\log_b a = x \iff b^x=a.$$ Thus, the logarithmic function is the inverse of the exponential function. When the base is $e$, it is called the natural logarithm, written as $$\ln x \quad \text{or} \quad \log_e x.$$

For $y=\ln x$:

• Domain: $(0,\infty)$
• Range: $\mathbb{R}$

Key properties

• It always passes through $(1,0)$ because $\ln 1=0$.
• It is defined only for positive real numbers.
• It is strictly increasing.
• It is the inverse of $e^x$.

The graphs of $y=e^x$ and $y=\ln x$ are reflections of each other in the line $y=x$.

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Properties of Logarithms

Let $x,y>0$ and let the base $b$ satisfy $b>0,\ b\neq 1$. Then:

1. Product rule $$\log_b(xy)=\log_b x+\log_b y$$

2. Quotient rule $$\log_b\left(\frac{x}{y}\right)=\log_b x-\log_b y$$

3. Power rule $$\log_b(x^n)=n\log_b x$$

4. Change of base formula $$\log_a p = \frac{\log_b p}{\log_b a}$$ In particular, $$\log_a x = \frac{\ln x}{\ln a}.$$

5. Inverse relation with exponentials $$e^{\ln x}=x \quad (x>0), \qquad \ln(e^x)=x.$$

These identities are extremely important in differentiation, especially in logarithmic differentiation.

Important caution: $$\log(A+B) \ne \log A + \log B.$$ This is one of the most common mistakes students make.

Note: In many calculus contexts, when the base is not specified, $\log x$ is taken to mean the natural logarithm $\ln x$. In school-level problems, follow the convention being used in the chapter or textbook.

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Derivatives of Exponential and Logarithmic Functions

The standard derivative formulas are:

1. Derivative of natural exponential function $$\frac{d}{dx}(e^x)=e^x$$

2. Derivative of natural logarithm $$\frac{d}{dx}(\ln x)=\frac{1}{x}, \qquad x>0$$

3. Derivative of general exponential function $$\frac{d}{dx}(a^x)=a^x\ln a, \qquad a>0,\ a\ne 1$$

4. Derivative of logarithm to base $a$ $$\frac{d}{dx}(\log_a x)=\frac{1}{x\ln a}, \qquad x>0$$

These formulas are often used together with the chain rule, product rule, and quotient rule. For example, $$\frac{d}{dx}(e^{u(x)})=e^{u(x)}u'(x),$$ and $$\frac{d}{dx}(\ln(u(x)))=\frac{u'(x)}{u(x)}$$ whenever $u(x)>0$.

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Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used when ordinary differentiation becomes long or awkward.

It is especially useful in the following situations:

• when the function is of the form $$y=[u(x)]^{v(x)},$$ that is, a variable base raised to a variable exponent;
• when the function is a product or quotient of many complicated factors;
• when fractional powers, radicals, and multiple terms appear together in a form that can be simplified using logarithms.

Steps of logarithmic differentiation

Step 1: Let $$y=f(x).$$ Take natural logarithm on both sides: $$\ln y = \ln(f(x)).$$

Step 2: Use logarithmic properties to simplify the right-hand side. Convert:

• products into sums,
• quotients into differences,
• powers into multipliers.

Step 3: Differentiate both sides implicitly with respect to $x$. Since $$\frac{d}{dx}(\ln y)=\frac{1}{y}\frac{dy}{dx},$$ this produces an equation involving $\dfrac{dy}{dx}$.

Step 4: Solve for $\dfrac{dy}{dx}$.

Step 5: Replace $y$ by its original expression so that the final answer is written purely in terms of $x$.

Important caution: Logarithmic differentiation cannot be applied directly to a sum such as $$y=u+v$$ by writing $\ln y = \ln u + \ln v$, because that is false. In such cases, first split the function into separate terms and differentiate each part individually.

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Solved Examples

Example 1: Chain Rule with Exponential Function

Differentiate $y=e^{-x}$ with respect to $x$.

Solution: The function is a composition of the exponential function and the linear function $-x$. Let $$u(x)=-x.$$ Then $$y=e^{u(x)}.$$ Using the chain rule, $$\frac{dy}{dx}=e^{u(x)}\cdot u'(x).$$ Now, $$u'(x)=\frac{d}{dx}(-x)=-1.$$ Therefore, $$\frac{dy}{dx}=e^{-x}(-1)=-e^{-x}.$$

Answer: $$-e^{-x}$$

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Example 2: Quotient Rule with Exponential Function

Find the derivative of $y=\dfrac{e^x}{\sin x}$.

Solution: Let $$u=e^x, \qquad v=\sin x.$$ Then $$y=\frac{u}{v}.$$ Using the quotient rule, $$\frac{dy}{dx}=\frac{u'v-uv'}{v^2}.$$ Now, $$u'=e^x, \qquad v'=\cos x.$$ So, $$\frac{dy}{dx}=\frac{e^x\sin x - e^x\cos x}{\sin^2 x}.$$ Factor out $e^x$ in the numerator: $$\frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{\sin^2 x}.$$

Answer: $$\frac{e^x(\sin x-\cos x)}{\sin^2 x}$$

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Example 3: Chain Rule with Logarithmic Function

Differentiate $y=\log(\cos e^x)$ with respect to $x$.

Solution: Assuming $\log$ denotes natural logarithm, we write $$y=\ln(\cos e^x).$$ This is a composition of three functions:

• outermost: $\ln u$,
• middle: $\cos v$,
• innermost: $e^x$.

Using the chain rule, $$\frac{dy}{dx}=\frac{1}{\cos e^x}\cdot \frac{d}{dx}(\cos e^x).$$ Now, $$\frac{d}{dx}(\cos e^x)=-\sin(e^x)\cdot \frac{d}{dx}(e^x).$$ Since $$\frac{d}{dx}(e^x)=e^x,$$ we get $$\frac{d}{dx}(\cos e^x)=-\sin(e^x)e^x.$$ Substitute back: $$\frac{dy}{dx}=\frac{-e^x\sin(e^x)}{\cos(e^x)}=-e^x\tan(e^x).$$

Answer: $$-e^x\tan(e^x)$$

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Example 4: Basic Logarithmic Differentiation

Differentiate $y=x^x$ with respect to $x$.

Solution: Since both the base and exponent involve the variable $x$, ordinary power rules do not apply directly. So we use logarithmic differentiation.

Take natural logarithm on both sides: $$\ln y = \ln(x^x).$$ Using the logarithmic power rule, $$\ln y = x\ln x.$$ Now differentiate both sides with respect to $x$. On the left, $$\frac{d}{dx}(\ln y)=\frac{1}{y}\frac{dy}{dx}.$$ On the right, use the product rule: $$\frac{d}{dx}(x\ln x)=x\cdot\frac{1}{x}+\ln x\cdot 1 = 1+\ln x.$$ So, $$\frac{1}{y}\frac{dy}{dx}=1+\ln x.$$ Multiply both sides by $y$: $$\frac{dy}{dx}=y(1+\ln x).$$ Now substitute $y=x^x$: $$\frac{dy}{dx}=x^x(1+\ln x).$$

Answer: $$x^x(1+\ln x)$$

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Example 5: Logarithmic Differentiation for Complex Products

Differentiate $y=\sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ with respect to $x$.

Solution: First rewrite the square root as power $\tfrac12$: $$y=\left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{1/2}.$$ Take natural logarithm on both sides: $$\ln y = \frac12 \ln\left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right).$$ Now expand using logarithmic rules: $$\ln y = \frac12\big[\ln(x-1)+\ln(x-2)-\ln(x-3)-\ln(x-4)-\ln(x-5)\big].$$ Differentiate both sides: $$\frac{1}{y}\frac{dy}{dx} = \frac12\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right].$$ Multiply through by $y$: $$\frac{dy}{dx} = \frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right].$$ Finally substitute the original value of $y$ back: $$\frac{dy}{dx}=\frac12\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right].$$

Answer: $$\frac12\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]$$

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Example 6: Warning - Sum of Variable Powers

Differentiate $y=x^{\sin x}+(\sin x)^{\cos x}$.

Solution: A logarithm cannot be taken directly over a sum in the form $$\ln(A+B)=\ln A + \ln B,$$ which is false. So we must differentiate the two terms separately.

Let $$u=x^{\sin x}, \qquad v=(\sin x)^{\cos x}.$$ Then $$y=u+v \implies \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}.$$

First term: $u=x^{\sin x}$ Take logarithm: $$\ln u = \sin x\ln x.$$ Differentiate: $$\frac{1}{u}\frac{du}{dx}=\cos x\ln x + \sin x\cdot\frac{1}{x}.$$ So, $$\frac{du}{dx}=x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right).$$

Second term: $v=(\sin x)^{\cos x}$ Take logarithm: $$\ln v = \cos x\ln(\sin x).$$ Differentiate using product rule: $$\frac{1}{v}\frac{dv}{dx}=(-\sin x)\ln(\sin x) + \cos x\cdot\frac{\cos x}{\sin x}.$$ Since $$\frac{\cos x}{\sin x}=\cot x,$$ this becomes $$\frac{1}{v}\frac{dv}{dx} = -\sin x\ln(\sin x) + \cos x\cot x.$$ Hence, $$\frac{dv}{dx}=(\sin x)^{\cos x}\big(\cos x\cot x - \sin x\ln(\sin x)\big).$$

Add both parts: $$\frac{dy}{dx}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\ln x\right) + (\sin x)^{\cos x}\big(\cos x\cot x - \sin x\ln(\sin x)\big).$$

Answer: $$x^{\sin x}\left(\frac{\sin x}{x}+\cos x\ln x\right) + (\sin x)^{\cos x}\big(\cos x\cot x - \sin x\ln(\sin x)\big)$$

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Example 7: Implicit Equation with Logarithmic Differentiation

Find $\frac{dy}{dx}$ if $y^x=x^y$.

Solution: Take natural logarithm on both sides: $$\ln(y^x)=\ln(x^y).$$ Using power rule of logarithms, $$x\ln y = y\ln x.$$ Now differentiate both sides implicitly with respect to $x$. On the left, using the product rule: $$\frac{d}{dx}(x\ln y)=\ln y + x\cdot\frac{1}{y}\frac{dy}{dx}.$$ On the right: $$\frac{d}{dx}(y\ln x)=\frac{dy}{dx}\ln x + y\cdot\frac{1}{x}.$$ So, $$\ln y + \frac{x}{y}\frac{dy}{dx} = \ln x\frac{dy}{dx} + \frac{y}{x}.$$ Bring derivative terms to one side: $$\frac{x}{y}\frac{dy}{dx} - \ln x\frac{dy}{dx} = \frac{y}{x} - \ln y.$$ Factor out $\dfrac{dy}{dx}$: $$\frac{dy}{dx}\left(\frac{x}{y}-\ln x\right)=\frac{y}{x}-\ln y.$$ Hence, $$\frac{dy}{dx}=\frac{\frac{y}{x}-\ln y}{\frac{x}{y}-\ln x}.$$ Multiplying numerator and denominator suitably to remove fractions, $$\frac{dy}{dx}=\frac{y(y-x\ln y)}{x(x-y\ln x)}.$$

Answer: $$\frac{y(y-x\ln y)}{x(x-y\ln x)}$$

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Example 8: Derivative of a Function with Respect to Another Base

Differentiate $y=a^x$ with respect to $x$, where $a$ is a positive constant, using logarithmic differentiation.

Solution: Take natural logarithm of both sides: $$\ln y = \ln(a^x)=x\ln a.$$ Differentiate both sides: $$\frac{1}{y}\frac{dy}{dx}=\ln a.$$ Now multiply by $y$: $$\frac{dy}{dx}=y\ln a.$$ Substitute $y=a^x$: $$\frac{dy}{dx}=a^x\ln a.$$

Answer: $$a^x\ln a$$

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Example 9: Change of Base Formula

Find the derivative of $y=\log_7(\log x)$.

Solution: Use the change-of-base formula: $$\log_7(\log x)=\frac{\ln(\log x)}{\ln 7}.$$ So, $$y=\frac{1}{\ln 7}\ln(\log x).$$ Treat $\dfrac{1}{\ln 7}$ as a constant and differentiate: $$\frac{dy}{dx}=\frac{1}{\ln 7}\cdot\frac{d}{dx}[\ln(\log x)].$$ Now use the chain rule: $$\frac{d}{dx}[\ln(\log x)] = \frac{1}{\log x}\cdot\frac{d}{dx}(\log x).$$ Assuming $\log x=\ln x$, we get $$\frac{d}{dx}(\log x)=\frac{1}{x}.$$ Therefore, $$\frac{dy}{dx}=\frac{1}{\ln 7}\cdot\frac{1}{\log x}\cdot\frac{1}{x} =\frac{1}{x\ln 7\,\log x}.$$

Answer: $$\frac{1}{x\ln 7\,\log x}$$

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Example 10: Inverse Trigonometry inside Exponents

Differentiate $y=e^{\sin^{-1}x}$.

Solution: This is a composition of the exponential function and inverse sine function. Using the chain rule, $$\frac{dy}{dx}=e^{\sin^{-1}x}\cdot\frac{d}{dx}(\sin^{-1}x).$$ Now use the standard derivative formula: $$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}.$$ Therefore, $$\frac{dy}{dx}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}.$$

Answer: $$\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}$$

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Example 11: Complex Logarithmic Differentiation

Find $\frac{dy}{dx}$ if $y=x^{\frac{1}{x}}$.

Solution: Since both base and exponent involve $x$, use logarithmic differentiation. Take natural logarithm: $$\ln y = \ln\left(x^{1/x}\right)=\frac{1}{x}\ln x.$$ Differentiate both sides: $$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{\ln x}{x}\right).$$ Use either product rule or quotient rule. Writing as $x^{-1}\ln x$, $$\frac{d}{dx}(x^{-1}\ln x)=x^{-1}\cdot\frac{1}{x} + \ln x\cdot(-x^{-2}).$$ So, $$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{\ln x}{x^2}=\frac{1-\ln x}{x^2}.$$ Multiply through by $y$: $$\frac{dy}{dx}=y\cdot\frac{1-\ln x}{x^2}.$$ Now substitute $y=x^{1/x}$: $$\frac{dy}{dx}=x^{1/x}\cdot\frac{1-\ln x}{x^2}.$$

Answer: $$x^{1/x}\left(\frac{1-\ln x}{x^2}\right)$$