Not every relation is handed to us as $y=f(x)$. Three techniques — implicit, logarithmic and parametric differentiation — extend the machinery to relations where $y$ is tangled with $x$, raised to a variable power, or expressed through a parameter. The chapter closes with two existence theorems about derivatives on an interval.
Implicit differentiation
When $x$ and $y$ are bound together in one equation (e.g. $x^2+y^2=25$, or $\sin(xy)=x$), we cannot always solve for $y$. Instead, differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every time a $y$ is differentiated, the chain rule attaches a factor $\dfrac{dy}{dx}$:
$$\frac{d}{dx}\big(y^{n}\big)=n\,y^{n-1}\frac{dy}{dx},\qquad \frac{d}{dx}\big(\sin y\big)=\cos y\,\frac{dy}{dx}.$$
After differentiating, collect all the $\dfrac{dy}{dx}$ terms on one side and solve algebraically. The answer usually involves both $x$ and $y$, which is perfectly acceptable.
Logarithmic differentiation
Two situations call for taking the logarithm of both sides first:
- A variable raised to a variable power, such as $y=x^{x}$ or $y=(\sin x)^{\cos x}$. Here neither the power rule (the exponent is not constant) nor the exponential rule (the base is not constant) applies directly. Taking $\log$ turns the power into a product: $\log y=v(x)\log u(x)$, which the product rule then handles. For $y=u(x)^{v(x)}$ this is the only elementary route.
- A long product or quotient of many factors, such as $y=\dfrac{(x-1)(x-2)}{(x-3)(x-4)}$. Taking $\log$ converts products into sums and quotients into differences of logs, which differentiate term by term — far cleaner than repeated product/quotient rules.
The mechanical step is always the same: after $\log y=\dots$, differentiate to get $\dfrac{1}{y}\dfrac{dy}{dx}=\dots$, then multiply through by $y$ (re-substituting the original expression).
Parametric differentiation
If a curve is given through a parameter $t$ as $x=f(t)$ and $y=g(t)$, then by the chain rule
$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{g'(t)}{f'(t)},\qquad f'(t)\ne 0.$$
For the second derivative, do not simply divide twice — you must differentiate $\dfrac{dy}{dx}$ (itself a function of $t$) with respect to $x$, which means dividing by $\dfrac{dx}{dt}$ once more:
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\!\left(\frac{dy}{dx}\right)=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}.$$
Rolle's Theorem
If $f$ is (i) continuous on $[a,b]$, (ii) differentiable on $(a,b)$, and (iii) $f(a)=f(b)$, then there exists at least one $c\in(a,b)$ with
$$f'(c)=0.$$
Geometrically: a smooth curve that starts and ends at the same height must have at least one point with a horizontal tangent (a turning point) in between. All three hypotheses are essential — $f(x)=|x|$ on $[-1,1]$ satisfies (i) and (iii) but fails (ii) at $0$, and indeed has no horizontal tangent.
Lagrange's Mean Value Theorem (MVT)
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c\in(a,b)$ with
$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$
Geometrically: somewhere inside the interval the tangent is parallel to the chord joining the endpoints — the instantaneous rate of change equals the average rate of change at least once. Rolle's theorem is the special case $f(a)=f(b)$, where the chord is horizontal and $f'(c)=0$. To verify the MVT for a given $f$, check the two hypotheses, compute the chord slope, then solve $f'(c)=\text{slope}$ for $c$ and confirm it lies in $(a,b)$.
Find $\dfrac{dy}{dx}$ if $x^2+y^2=25$.
SolutionDifferentiate implicitly: $2x+2y\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}.$
Answer: $\dfrac{dy}{dx}=-\dfrac{x}{y}.$
Differentiate $y=x^{x}$ ($x>0$).
SolutionTake logs: $\log y=x\log x$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx}=\log x+1$. So $\dfrac{dy}{dx}=x^{x}(1+\log x).$
Answer: $\dfrac{dy}{dx}=x^{x}(1+\log x).$
If $x=at^2,\ y=2at$, find $\dfrac{dy}{dx}$.
Solution$\dfrac{dx}{dt}=2at,\ \dfrac{dy}{dt}=2a$. Hence $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}.$
Answer: $\dfrac{dy}{dx}=\dfrac{1}{t}.$
Find $\dfrac{d^2y}{dx^2}$ for $y=x^3$.
Solution$\dfrac{dy}{dx}=3x^2$, then $\dfrac{d^2y}{dx^2}=6x.$
Answer: $\dfrac{d^2y}{dx^2}=6x.$
Find $\dfrac{dy}{dx}$ if $\sin(xy)+\dfrac{x}{y}=x^2$.
SolutionDifferentiate term by term, treating $y$ as a function of $x$. For $\sin(xy)$: $\cos(xy)\big(y+x\,y'\big)$. For $\dfrac{x}{y}$ (quotient): $\dfrac{y-x y'}{y^2}$. The right side gives $2x$. So $\cos(xy)\,(y+xy')+\dfrac{y-xy'}{y^2}=2x.$ Collect $y'$: $y'\Big(x\cos(xy)-\dfrac{x}{y^2}\Big)=2x-y\cos(xy)-\dfrac{1}{y}$, hence $\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)-\tfrac{1}{y}}{x\cos(xy)-\tfrac{x}{y^2}}.$
Answer: $\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)-\tfrac{1}{y}}{x\cos(xy)-\tfrac{x}{y^2}}.$
Differentiate $y=(\sin x)^{\cos x}$.
SolutionTake logs: $\log y=\cos x\,\log(\sin x)$. Differentiate (product rule on the right): $\dfrac{1}{y}\dfrac{dy}{dx}=-\sin x\,\log(\sin x)+\cos x\cdot\dfrac{\cos x}{\sin x}.$ Multiply by $y=(\sin x)^{\cos x}$: $\dfrac{dy}{dx}=(\sin x)^{\cos x}\Big(\cos x\cot x-\sin x\,\log(\sin x)\Big).$
Answer: $\dfrac{dy}{dx}=(\sin x)^{\cos x}\Big(\cos x\cot x-\sin x\,\log(\sin x)\Big).$
Differentiate $y=\dfrac{(x-1)(x-2)}{(x-3)(x-4)}$ using logarithms.
Solution$\log y=\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}.$ Therefore $\dfrac{dy}{dx}=\dfrac{(x-1)(x-2)}{(x-3)(x-4)}\left[\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}\right].$
Answer: $\dfrac{dy}{dx}=\dfrac{(x-1)(x-2)}{(x-3)(x-4)}\left[\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}\right].$
If $x=a(\theta-\sin\theta)$ and $y=a(1-\cos\theta)$ (a cycloid), find $\dfrac{dy}{dx}$.
Solution$\dfrac{dx}{d\theta}=a(1-\cos\theta)$ and $\dfrac{dy}{d\theta}=a\sin\theta$. So $\dfrac{dy}{dx}=\dfrac{a\sin\theta}{a(1-\cos\theta)}=\dfrac{\sin\theta}{1-\cos\theta}.$ Using $\sin\theta=2\sin\tfrac\theta2\cos\tfrac\theta2$ and $1-\cos\theta=2\sin^2\tfrac\theta2$, this simplifies to $\cot\dfrac{\theta}{2}.$
Answer: $\dfrac{dy}{dx}=\cot\dfrac{\theta}{2}.$
Verify Rolle's theorem for $f(x)=x^2-4x+3$ on $[1,3]$ and find $c$.
Solution$f$ is a polynomial, so continuous on $[1,3]$ and differentiable on $(1,3)$. Endpoints: $f(1)=1-4+3=0$ and $f(3)=9-12+3=0$, so $f(1)=f(3)$ — all three hypotheses hold. Now $f'(x)=2x-4$; setting $f'(c)=0$ gives $c=2$, which lies in $(1,3)$. Rolle's theorem is verified, with $c=2$.
Answer: Rolle's theorem is verified, with $c=2$.
Verify Lagrange's Mean Value Theorem for $f(x)=\sqrt{x}$ on $[1,4]$ and find $c$.
Solution$f$ is continuous on $[1,4]$ and differentiable on $(1,4)$. Chord slope $=\dfrac{f(4)-f(1)}{4-1}=\dfrac{2-1}{3}=\dfrac{1}{3}.$ Since $f'(x)=\dfrac{1}{2\sqrt{x}}$, set $\dfrac{1}{2\sqrt{c}}=\dfrac{1}{3}\Rightarrow \sqrt{c}=\dfrac{3}{2}\Rightarrow c=\dfrac{9}{4}.$ As $\tfrac94=2.25\in(1,4)$, the MVT is verified with $c=\dfrac{9}{4}.$
Answer: MVT is verified with $c=\dfrac{9}{4}.$
If $x^{y}=e^{x-y}$, prove that $\dfrac{dy}{dx}=\dfrac{\log x}{(1+\log x)^2}$.
SolutionTake logs: $y\log x=x-y$, so $y(1+\log x)=x$ and $y=\dfrac{x}{1+\log x}.$ Differentiate by the quotient rule: $\dfrac{dy}{dx}=\dfrac{(1+\log x)\cdot 1-x\cdot\tfrac1x}{(1+\log x)^2}=\dfrac{1+\log x-1}{(1+\log x)^2}=\dfrac{\log x}{(1+\log x)^2}.$
Answer: $\dfrac{dy}{dx}=\dfrac{\log x}{(1+\log x)^2}.$