Exercises — Derivatives of ln x and logₐ(x)

The constant $5$ has slope $0$. So $f^{\prime }(x)=\frac{1}{x}$. Why: Adding a constant shifts the graph up but never tilts it — the steepness is untouched.

Here the base is $a=10$, so $\ln a=\ln 10$. $g^{\prime }(x)=\frac{1}{x\ln 10}.$ Why not $1/x$? Only base $e$ gives clean $1/x$, because $\ln e=1$. Any other base pays the factor $\frac{1}{\ln a}$.

The constant $7$ rides along as a multiplier: $h^{\prime }(x)=7\cdot \frac{1}{x}=\frac{7}{x}.$

Chain rule: inside is $u=2x$, so $u^{\prime }=2$, giving $\frac{u^{\prime }}{u}=\frac{2}{2x}=\frac{1}{x}$. So (ii) and (iii) are the same thing — both correct; (i) is wrong. Deeper reason: $\ln (2x)=\ln 2+\ln x$, and $\ln 2$ is a constant that dies when differentiated. The multiplier inside a log never changes the derivative.

Inside $u=3x^2+1$, so $u^{\prime }=6x$. $f^{\prime }(x)=\frac{u^{\prime }}{u}=\frac{6x}{3x^2+1}.$

Inside $u=\sin x$, so $u^{\prime }=\cos x$. $g^{\prime }(x)=\frac{\cos x}{\sin x}=\cot x.$ Why: "derivative of inside over inside." The Chain Rule does all the work.

$h^{\prime }(x)=\frac{(5x{)}^{\prime }}{(5x)\ln 2}=\frac{5}{5x\ln 2}=\frac{1}{x\ln 2}.$ Why the $5$ vanishes: ${\log }_2(5x)={\log }_{2}5+{\log }_{2}x$; the constant ${\log }_{2}5$ has slope $0$.

Simplify first using Logarithm Laws: $\ln (x^2+1{)}^3=3\ln (x^2+1)$. $k^{\prime }(x)=3\cdot \frac{2x}{x^2+1}=\frac{6x}{x^2+1}.$ Why simplify: the power law $\ln A^n=n\ln A$ turns a scary chain into a one-liner.

Split with the quotient law: $\ln \frac{x}{x+1}=\ln x-\ln (x+1)$. $f^{\prime }(x)=\frac{1}{x}-\frac{1}{x+1}=\frac{(x+1)-x}{x(x+1)}=\frac{1}{x(x+1)}.$ Why split: differentiating two simple logs beats a quotient rule buried under a chain rule.

For $x>0$: $\ln |x|=\ln x$, derivative $\frac{1}{x}$. For $x<0$: $\ln |x|=\ln (-x)$; inside $u=-x$, $u^{\prime }=-1$, so $\frac{-1}{-x}=\frac{1}{x}$. Both branches give the same answer. See the figure: the two curve pieces are mirror images, and their slopes match the single formula $1/x$ everywhere except the forbidden point $x=0$. $\boxed{\frac{d}{dx}\ln |x|=\frac{1}{x},x\ne 0.}$

Inside $u=\cos x$, $u^{\prime }=-\sin x$. $g^{\prime }(x)=\frac{-\sin x}{\cos x}=-\tan x.$

Domain: need $x^2-4>0\Rightarrow |x|>2$, i.e. $x<-2$ or $x>2$. Derivative: inside $u=x^2-4$, $u^{\prime }=2x$: $\frac{2x}{x^2-4}.$ Note this is negative on $(-\infty ,-2)$ and positive on $(2,\infty )$ — the graph falls then rises, matching two separate branches.

The variable sits in the exponent, so no plain power rule applies. Take $\ln$ of both sides: $\ln y=x\ln x.$ Differentiate (left side by Implicit Differentiation, since $y$ depends on $x$): $\frac{y^{\prime }}{y}=\ln x+x\cdot \frac{1}{x}=\ln x+1.$ Multiply back by $y=x^x$: $y^{\prime }=x^x(\ln x+1).$

Take $\ln$ and expand with log laws (products → sums, powers → multipliers): $\ln y=2\ln (x+1)+\frac{1}{2}\ln (x-1)-4\ln (x+3).$ Differentiate term by term: $\frac{y^{\prime }}{y}=\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{4}{x+3}.$ So $y^{\prime }=\frac{(x+1{)}^2\sqrt{x-1}}{(x+3{)}^4}\left[\frac{2}{x+1}+\frac{1}{2(x-1)}-\frac{4}{x+3}\right].$ Why this beats quotient+product+chain: logs convert one tangled fraction into a clean sum of easy pieces.

Variable base and variable exponent — logarithmic differentiation is forced. $\ln y=\sin x\cdot \ln x.$ $\frac{y^{\prime }}{y}=\cos x\cdot \ln x+\sin x\cdot \frac{1}{x}.$ $y^{\prime }=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right).$

Change of base with the new base being $x$: $y={\log }_{x}2=\frac{\ln 2}{\ln x}=\ln 2\cdot (\ln x{)}^{-1}.$ $\ln 2$ is a constant. Differentiate the power: $y^{\prime }=\ln 2\cdot (-1)(\ln x{)}^{-2}\cdot \frac{1}{x}=-\frac{\ln 2}{x(\ln x{)}^2}.$ Why this is subtle: the variable is in the base of the log, so you cannot use $\frac{1}{x\ln a}$ directly — that formula assumes a constant base. Rewrite first.

$\frac{d}{dx}\ln x={\lim }_{h\to 0}\frac{\ln (x+h)-\ln x}{h}={\lim }_{h\to 0}\frac{1}{h}\ln \left(\frac{x+h}{x}\right).$ Why: the quotient law $\ln A-\ln B=\ln \frac{A}{B}$ collapses the difference. Substitute $t=h/x$ (so $h=xt$, and $t\to 0$ as $h\to 0$): $={\lim }_{t\to 0}\frac{1}{xt}\ln (1+t)=\frac{1}{x}{\lim }_{t\to 0}\frac{\ln (1+t)}{t}=\frac{1}{x}\cdot 1=\frac{1}{x}.$ The engine is the standard limit $\underset{t\to 0}{\lim }\frac{\ln (1+t)}{t}=1$, which itself follows from $(1+t{)}^{1/t}\to e$ and $\ln e=1$.

Point: at $x=e$, $y=\ln e=1$, so $(e,1)$. Slope: $y^{\prime }=\frac{1}{x}$, at $x=e$ gives $\frac{1}{e}$. Tangent: $y-1=\frac{1}{e}(x-e)\Rightarrow y=\frac{x}{e}-1+1=\frac{x}{e}$. $\boxed{y=\frac{x}{e}.}$ The tangent at $x=e$ passes through the origin — a clean, memorable fact. See figure.

$y^{\prime }=\frac{1}{x}=x^{-1}$, so $y^{\prime \prime }=-x^{-2}=-\frac{1}{x^2}$. For any real $x\ne 0$, $x^2>0$, hence $y^{\prime \prime }<0$ for all $x>0$. Interpretation: a negative second derivative means the curve is concave down everywhere — the slope $1/x$ keeps decreasing as $x$ grows. The hill of $\ln x$ never stops flattening.

Take $\ln$: $\ln y=v\ln u$. Differentiate both sides (product rule on the right, implicit on the left): $\frac{y^{\prime }}{y}=v^{\prime }\ln u+v\cdot \frac{u^{\prime }}{u}.$ Multiply by $y=u^v$: $\boxed{y^{\prime }=u^v\left(v^{\prime }\ln u+\frac{v{u}^{\prime }}{u}\right).}$ Check L4.1 ($u=x,v=x$): $u^{\prime }=1,v^{\prime }=1$, giving $x^x(\ln x+\frac{x}{x})=x^x(\ln x+1)$. ✓ Check L4.3 ($u=x,v=\sin x$): $u^{\prime }=1,v^{\prime }=\cos x$, giving $x^{\sin x}(\cos x\ln x+\frac{\sin x}{x})$. ✓

Derivative of ${\log }_{10}x$ ::: $\frac{1}{x\ln 10}$ $\frac{d}{dx}\ln (\sin x)$ ::: $\cot x$ $\frac{d}{dx}\ln |x|$ and where ::: $\frac{1}{x}$ for all $x\ne 0$ Tangent to $\ln x$ at $x=e$ ::: $y=x/e$ (through the origin) Second derivative of $\ln x$ ::: $-\frac{1}{x^2}$ (always concave down for $x>0$) Master rule for $u^v$ ::: $u^v\left(v^{\prime }\ln u+\frac{v{u}^{\prime }}{u}\right)$