5.4 Contents

1. Arithmetic Properties

2. Product Rule

3. Quotient Rule

4. Power Rule

5. Mean Value Theorem for Derivatives

6. Implicit Differentiation (Chain Rule)

7. Derivative of an Inverse Function

8. L'Hôpital's Rule

9. $d/dx\thinspace e^x$

1. Derivative of Logarithmic Functions 🔧

2. Derivative of Exponential Functions

3. $d/dx\thinspace x^x$

5.4.1 Arithmetic Properties

Constant Rule

$$\frac{d}{dx}(C)=0$$

Constant Multiple Rule

$$\frac{d}{dx}(C\cdot f(x))=C\cdot \frac{d}{dx}f(x)$$

Sum & Difference Rules

$$\frac{d}{dx}\big(f(x)\pm g(x)\big)=f'(x)\pm g'(x)$$

5.4.2 Product Rule

$$\frac{d}{dx}\big(f(x)\cdot g(x)\big)=f'(x)\cdot g(x)+f(x)\cdot g'(x)$$

Proof

Insert the product expression into the derivative function $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{\Delta x}$$ Add $f(x+\Delta x)\cdot g(x)$ minus itself into the numerator $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)\cdot g(x+\Delta x)+f(x+\Delta x)\cdot g(x)-f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)}{\Delta x}$$ Separate the fraction and apply the limit of sums property $$\lim\limits_{\Delta x\to 0}\bigg(\frac{f(x+\Delta x)\cdot g(x+\Delta x)-f(x+\Delta x)\cdot g(x)}{\Delta x}\bigg)+\lim\limits_{\Delta x\to 0}\bigg(\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)}{\Delta x}\bigg)$$ Factor each term to isolate derivative functions $$\lim\limits_{\Delta x\to 0}\bigg(f(x+\Delta x)\cdot\frac{g(x+\Delta x)-g(x)}{\Delta x}\bigg)+\lim\limits_{\Delta x\to 0}\bigg(g(x)\cdot\frac{f(x+\Delta x)-f(x)}{\Delta x}\bigg)$$ Rewrite in terms of derivatives $$\bigg(\lim\limits_{\Delta x\to 0}f(x+\Delta x)\bigg)\cdot\bigg(\frac{d}{dx}g(x)\bigg)+\bigg(\lim\limits_{\Delta x\to 0}g(x)\bigg)\cdot\bigg(\frac{d}{dx}g(x)\bigg)$$ Solve the limits

5.4.3 Quotient Rule

$$\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{g(x)^2}$$

Proof

Insert the quotient expression into the derivative function $$\lim\limits_{\Delta x\to 0}\frac{1}{\Delta x}\cdot\bigg(\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}\bigg)$$ Factor $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x+\Delta x)}{g(x)\cdot g(x+\Delta x)}$$ Add $f(x)\cdot g(x)$ minus itself into the numerator $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)\cdot g(x)+f(x)\cdot g(x)-f(x)\cdot g(x)-f(x)\cdot g(x+\Delta x)}{g(x)\cdot g(x+\Delta x)}$$ Separate the fraction and apply the limit of sums property $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)\cdot g(x)-f(x)\cdot g(x)}{g(x)\cdot g(x+\Delta x) \cdot\Delta x}+\lim\limits_{\Delta x\to 0}\frac{f(x)\cdot g(x)-f(x)\cdot g(x+\Delta x)}{g(x)\cdot g(x+\Delta x)\cdot\Delta x}$$ Cancel $g(x)$ in the first term $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{g(x+\Delta x) \cdot\Delta x}+\lim\limits_{\Delta x\to 0}\frac{f(x)\cdot g(x)-f(x)\cdot g(x+\Delta x)}{g(x)\cdot g(x+\Delta x)\cdot\Delta x}$$ Factor a negative from the second term $$\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{g(x+\Delta x) \cdot\Delta x}-\lim\limits_{\Delta x\to 0}\frac{f(x)\cdot g(x+\Delta x)-f(x)\cdot g(x)}{g(x)\cdot g(x+\Delta x)\cdot\Delta x}$$ Factor each term to isolate derivative functions $$\lim\limits_{\Delta x\to 0}\bigg(\frac{1}{g(x+\Delta x)}\cdot\frac{f(x+\Delta x)-f(x)}{\Delta x}\bigg)-\lim\limits_{\Delta x\to 0}\bigg(\frac{f(x)}{g(x)\cdot g(x+\Delta x)}\cdot\frac{g(x+\Delta x)-g(x)}{\Delta x}\bigg)$$ Rewrite in terms of derivatives $$\lim\limits_{\Delta x\to 0}\bigg(\frac{1}{g(x+\Delta x)}\bigg)\cdot\bigg(\frac{d}{dx}f(x)\bigg)-\lim\limits_{\Delta x\to 0}\bigg(\frac{f(x)}{g(x)\cdot g(x+\Delta x)}\bigg)\cdot\bigg(\frac{d}{dx}g(x)\bigg)$$ Solve the limits $$\frac{1}{g(x)}\cdot\frac{d}{dx}f(x)-\frac{f(x)}{g(x)^2}\cdot\frac{d}{dx}g(x)$$ Factor

5.4.4 Power Rule

$$\frac{d}{dx}x^n=n\cdot x^{n-1},\medspace \forall x\isin\R$$

Proof

Insert the exponential expression into the derivative function $$\frac{d}{dx}x^n=\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^n-x^n}{\Delta x}$$ Apply binomial expansion to the numerator $$\frac{d}{dx}x^n=\lim\limits_{\Delta x\to 0}\Bigg(\frac{1}{\Delta x}\cdot\binom{n}{1}\cdot x^{n-1}\cdot\Delta x+\binom{n}{2}\cdot x^{n-2}\cdot\Delta x^2+...+\binom{n}{n}\cdot\Delta x^n\Bigg)$$ Simplify $$\frac{d}{dx}x^n=\lim\limits_{\Delta x\to 0}\Bigg(\binom{n}{1}\cdot x^{n-1}+\binom{n}{2}\cdot x^{n-2}\cdot\Delta x+...+\Delta x^{n-1}\Bigg)$$ Solve the limit $$\frac{d}{dx}x^n=\binom{n}{1}\cdot x^{n-1}$$ Calculate the binomial

5.4.5 Mean Value Theorem for Derivatives

Within $[a,b]$, there exists a point $c$ such that $f'(c)$ is equal to the function's average rate of change $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

5.4.6 Implicit Differentiation (Chain Rule)

$$\frac{d}{dx}f\big(g(x)\big)=f'\big(g(x)\big)\cdot g'(x)$$ $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$

Deductive Logic

When comparing two rates, one is dependent on the other, and therefore dependent on the variable of the other, producing a scaling effect from the original function. Multiplying the two, like in percentages of percentages, yields the total change.

For example, in his 1985 book Calculus with Analytic Geometry, George Simmons writes, "If a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man."

5.4.7 Derivative of an Inverse Function

$$\big(f^{-1}\big)'\big(f(x)\big)=\frac{1}{f'(x)},\medspace f'(x)\ne 0$$

Proof

Using Langrange notation for $y=f(x)$, insert the inverse expression into the derivative function $$(f^{-1})'(y)=\lim\limits_{\Delta y\to 0}\frac{f^{-1}(y+\Delta y)-f^{-1}(y)}{\Delta y}$$ Factor $y-y$ into the denominator $$(f^{-1})'(y)=\lim\limits_{\Delta y\to 0}\frac{f^{-1}(y+\Delta y)-f^{-1}(y)}{y+\Delta y-y}$$ $y=f(x)\therefore f^{-1}(y)=x$. Substitute the inverse functions and $y$ values in terms of $x$. $$(f^{-1})'(y)=\lim\limits_{\Delta y\to 0}\frac{x+\Delta x-x}{f(x+\Delta x)-f(x)}$$ Cancel like terms $$(f^{-1})'(y)=\lim\limits_{\Delta y\to 0}\frac{\Delta x}{f(x+\Delta x)-f(x)}$$ Apply the reciprocal rule of division $$(f^{-1})'(y)=1\bigg/\lim\limits_{\Delta y\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ Rewrite the derivative function in terms of its differential

5.4.8 L'Hôpital's Rule

$$\text{If }\lim\limits_{x\to a}f(x)=\text{If }\lim\limits_{x\to a}g(x)=(0\text{ or }\pm\infin)$$ $$\text{then }\lim\limits_{x\to a}\frac{f(x)}{g(x)}=\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}$$

The indeterminate form $0\cdot\infin$

The rule holds when applying the reciprocal rule of division; $x\cdot y\to y/(1/x)$, also taking into account that 0 becomes ∞ or ∞ becomes 0

The indeterminate form $\infin-\infin$

The rule holds when factoring terms into one fraction, given that ∞ is the result of division by 0

Exponential Indeterminate Forms

In the cases of $\lim_{x\to a}f(x)^{g(x)}=(1^\infin\medspace\lor\medspace 0^0\medspace\lor\medspace\infin^0)$, the natural logarithm can be taken and equated with another limit $L$, then antilogged $$\lim\limits_{x\to a}g(x)\cdot\ln\big(f(x)\big)=\lim\limits_{x\to a}\frac{\ln\big(f(x)\big)}{1/g(x)}=\lim\limits_{x\to a}\frac{g(x)}{1/\ln\big(f(x)\big)}=L$$ $$\lim\limits_{x\to a}f(x)^{g(x)}=e^L$$

Proof

Lef $f'(x)$ and $g'(x)$ be continuous at $a$, and $f(a)=g(a)=0$ so that $$\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}=\frac{f'(x)}{g'(x)}$$ Apply the mean value theorem for derivatives $$\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to a}\bigg(\frac{f(x)-f(a)}{x-a}\bigg)\bigg/\bigg(\frac{g(x)-g(a)}{x-a}\bigg)$$ Simplify $$\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}$$ Evaluate the limit $$\lim\limits_{x\to a}\frac{f'(x)}{g'(x)}=\frac{0}{0}=\frac{f(x)}{g(x)}$$ Beginning using the reciprocal rule of division yields $$\lim\limits_{x\to a}\frac{1/f'(x)}{1/g'(x)}=\frac{\infin}{\infin}=\frac{1/f(x)}{1/g(x)}$$

5.4.9 $d/dx\thinspace e^x$

$$\frac{d}{dx}e^x=e^x$$

Proof

Insert $e^x$ into the derivative function $$\lim\limits_{\Delta x\to 0}\frac{e^{x+\Delta x}-e^x}{\Delta x}$$ Factor the exponential power $$\lim\limits_{\Delta x\to 0}\frac{e^x\cdot e^{\Delta x}-e^x}{\Delta x}$$ Factor $e^x$ from the limit $$e^x\cdot\lim\limits_{\Delta x\to 0}\frac{e^{\Delta x}-1}{\Delta x}$$ Evaluate using the $e\thickspace$limit

5.4.A Derivative of Logarithmic Functions 🔧

$$\frac{d}{dx}=\log_b x=\big(x\cdot\ln(b)\big)^{-1}$$ $$\frac{d}{dx}\ln(x)=x^{-1}$$ $$x>0\medspace\land\medspace b\ne1$$

Proof

Given $$\frac{d}{dx}\ln(x)$$ Define components for the derivative of an inverse function $$f(x)=e^x\qquad f'(x)=e^x\qquad f^{-1}(x)=\ln(x)$$ Since $f\big(f^{-1}(x)\big)=e^{\ln(x)}$ $$\frac{d}{dx}\ln(x)=\frac{1}{e^{\ln(x)}}$$ Simplify $$\frac{d}{dx}\ln(x)=\frac{1}{x}$$ Given $$\frac{d}{dx}\log_b x$$ Apply the change of base rule with base $e$ $$\frac{d}{dx}\log_b x=\frac{d}{dx}\frac{\ln(x)}{\ln(b)}$$ Apply the constant rule $$\frac{d}{dx}\log_b x=\frac{1}{\ln(b)}\cdot\frac{d}{dx}\ln(x)$$ Solve for $d/dx\thinspace\ln(x)$

5.4.B Derivative of Exponential Functions

$$\frac{d}{dx}b^x=b^x\cdot\ln(b)$$

Proof

Let $$y=b^x$$ Take the natural logarithm $$\ln(y)=\ln(b^x)$$ Apply the logarithms power rule $$\ln(y)=x\cdot\ln(b)$$ Derive $$\frac{d}{dx}\ln(y)=\frac{d}{dx}x\cdot\ln(b)$$ Apply the chain rule to derive the natural logarithm $$y^{-1}\cdot\frac{dy}{dx}=\frac{d}{dx}x\cdot\ln(b)$$ Evaluate the derivative on the right $$y^{-1}\cdot\frac{dy}{dx}=\ln(b)$$ Multiply by $y$ $$\frac{dy}{dx}=y\cdot\ln(b)$$ Substitute $b^x$ for $y$

5.4.C $d/dx\thinspace x^x$

$$\frac{d}{dx}x^x=x^x\cdot \big(1+\ln(x)\big)$$

Proof

Let $$y=x^x$$ Take the natural logarithm $$\ln(y)=\ln(x^x)$$ Apply the logarithms power rule $$\ln(y)=x\cdot\ln(x)$$ Derive $$\frac{d}{dx}\ln(y)=\frac{d}{dx}x\cdot\ln(x)$$ Define components of the derivatives product rule $$ \begin{matrix} f(x)=x & g(x)=\ln(x) \\ f'(x)=1 & g'(x)=1/x \end{matrix} $$ Apply the chain rule to derive the natural logarithm, and derive the product on the right $$\frac{1}{y}\cdot\frac{dy}{dx}=x\cdot\frac{1}{x}+\ln(x)\cdot 1=1+\ln(x)$$ Multiply by $y$ $$\frac{dy}{dx}=y\cdot\big(1+\ln(x)\big)$$ Substitute $x^x$ for $y$