Derivatives of Exponential and Logarithmic Functions (Base ≠ e)

Derivative Rules for Bases $a \neq e$ - Exponential Rule

• For $f(x)=a^x$, $\displaystyle f'(x)= (\ln a)\,a^x$

• Chain-rule extension: if $f(x)=a^{u(x)}$ then $\displaystyle f'(x)= (\ln a)\,a^{u(x)}\,u'(x)$ - Logarithmic Rule

• For $g(x)=\log_a x$, $\displaystyle g'(x)=\frac1{(\ln a)\,x}$ • Chain-rule extension: if $g(x)=\log_a u(x)$ then $\displaystyle g'(x)=\frac{u'(x)}{(\ln a)\,u(x)}$ ### Proof Sketches - Derivative of $a^x$

• Rewrite: $a^x = e^{x\ln a}$ (log property).

• Differentiate using $\frac{d}{dx}e^{u}=e^{u}u'$:

$\frac{d}{dx}a^x = e^{x\ln a}(\ln a)=a^x\ln a$. - Derivative of $\log_a x$ • Change of base: $\log_a x = \frac{\ln x}{\ln a}$.

• Differentiate: $\frac{1}{\ln a}\,\frac{1}{x}$ (since $\ln a$ is constant wrt $x$).

• Result: $\frac1{(\ln a)\,x}$. ### Worked Examples #### Example 1\ Derivative of $f(x)=5^{x^3-4}$ - Identify parameters: $a=5,\;u=x^3-4,\;u'=3x^2$. - Apply chain rule form:

$f'(x)=(\ln 5)\,5^{x^3-4}\,(3x^2)=3(\ln 5)\,x^2\,5^{x^3-4}$. - Algebra tip: constants like $3\ln5$ usually front-loaded. #### Example 2\ Derivative of $g(x)=\log_3(2x^2-5x)$ - Parameters: $a=3,\;u=2x^2-5x,\;u'=4x-5$. - Rule: $g'(x)=\frac{u'}{(\ln a)\,u}$

$\displaystyle g'(x)=\frac{4x-5}{(\ln 3)\,(2x^2-5x)}$. - Factor note: denominator has a common factor $x$ but does not cancel with numerator. #### Example 3\ Product $h(x)=8^x\,\log_9 x$ - Write as $f(x)g(x)$ with

$f(x)=8^x,\;f'(x)=(\ln8)8^x$

$g(x)=\log_9 x,\;g'(x)=\frac1{(\ln9)x}$. - Product rule: $h' = f'g + fg'$

$h'(x)=(\ln8)8^x\log_9 x + 8^x\frac1{(\ln9)x}$. - Factor $8^x$:

$h'(x)=8^x\Big[(\ln8)\log_9 x + \frac1{(\ln9)x}\Big]$. - Log simplifications:

$\ln9 = \ln(3^2)=2\ln3$, $\ln8 = \ln(2^3)=3\ln2$.

Final compact form:

$h'(x)=8^x\Big[3(\ln2)\,\log_9 x + \frac1{2(\ln3)x}\Big]$. #### Example 4\ Quotient $q(x)=\dfrac{2^x}{3x-1}$ - $f(x)=2^x,\;f'(x)=(\ln2)2^x$

$g(x)=3x-1,\;g'(x)=3$. - Quotient rule: $q'=\frac{f'g-fg'}{g^2}$

$q'(x)=\frac{(\ln2)2^x(3x-1)-2^x(3)}{(3x-1)^2}$. - Factor $2^x$ in numerator:

$q'(x)=\frac{2^x[(\ln2)(3x-1)-3]}{(3x-1)^2}$. ### Application: Double-Declining Balance Depreciation #### Model - Purchase price $\$4,600$. - Salvage value function: $\displaystyle V(t)=4600\,(0.85)^t$ where $t$ = years. #### (a) Value After 2 Years - Evaluate: $V(2)=4600\,(0.85)^2 = \$3,325.50$. - Interpretation: machine is worth about $\$3,325.50$ after 2 years. #### (b) Instantaneous Rate of Change - Derivative:

$\displaystyle V'(t)=4600\,(\ln0.85)\,(0.85)^t$ (since $a=0.85$). #### (c) Rate After 3 Years - Evaluate: $V'(3)=4600\,(\ln0.85)\,(0.85)^3 \approx -\$459.11$ per year. - Interpretation: at $t=3$ the machine’s value is decreasing by about $\$459.11$ each year.

• Negative sign \Rightarrow decline.

• Rate is instantaneous—will change for $$t \neq 3$