Implicit Differentiation and Special Forms

For a relation $F(x,y)=0$ where $y$ is an (implicit) function of $x$:

• Differentiate both sides with respect to $x$.
• Every $y$-term picks up a factor $\dfrac{dy}{dx}$ (chain rule): $\dfrac{d}{dx}(y^n)=n y^{n-1}\dfrac{dy}{dx}$, $\dfrac{d}{dx}(xy)=y+x\dfrac{dy}{dx}$.
• Gather all $\dfrac{dy}{dx}$ terms on one side and solve for $\dfrac{dy}{dx}$.

The slope of the tangent at a point is $\dfrac{dy}{dx}$ evaluated there.

For a relation tying $x+y$ to a product or transcendental expression, evaluated at a point:

• Find the point. Substitute the given $x$ into the original equation to solve for $y$. For $\log(x+y)=2xy$ at $x=0$: $\log y = 0 \Rightarrow y = 1$.
• Differentiate implicitly, then substitute the full point $(x,y)$ and solve for $\dfrac{dy}{dx}$.

For $\log(x+y)=2xy$: $\dfrac{1+y'}{x+y} = 2(y + xy')$; at $(0,1)$ this gives $1+y' = 2y' \Rightarrow y' = 1$.

For a relation where $y$ appears in an exponent on one or both sides:

• Take $\log$ of both sides to bring exponents down: $\log(A^{B}) = B\log A$.
• Differentiate implicitly (product rule on each $B\log A$ term).
• Collect and solve for $\dfrac{dy}{dx}$.

Example shape: $(2x)^{2y}=4e^{2x-2y}$ becomes $2y\log(2x) = \log 4 + 2x - 2y$; differentiating yields $(1+\log 2x)^2\dfrac{dy}{dx} = x\log 2x - \log 2x$.

Given $\tan y = \dfrac{x\sin\alpha}{1 - x\cos\alpha}$:

• Differentiate: $\sec^2 y\,\dfrac{dy}{dx} = \dfrac{d}{dx}\!\left[\dfrac{x\sin\alpha}{1-x\cos\alpha}\right]$ (quotient rule on the right).
• Replace $\sec^2 y = 1 + \tan^2 y$ and substitute $\tan y$ from the given relation; the algebra collapses to

$\dfrac{dy}{dx} = \dfrac{\sin\alpha}{1 - 2x\cos\alpha + x^2}$. Matching to $\dfrac{m}{x^2 + 2nx + 1}$ gives $m = \sin\alpha$, $n = -\cos\alpha$, so $m^2 + n^2 = 1$.

Typical shape: $y^{1/m} + y^{-1/m} = 2x$ (or similar), prove $(x^2-1)(y')^2 = m^2 y^2$.

• Unwrap to explicit $y$. Set $t = y^{1/m}$; then $t + \dfrac{1}{t} = 2x$ is a quadratic $t^2 - 2xt + 1 = 0$, giving $t = x + \sqrt{x^2-1}$, so $y = \left(x+\sqrt{x^2-1}\right)^{m}$.
• Differentiate. $\dfrac{dy}{dx} = \dfrac{m y}{\sqrt{x^2-1}}$.
• Square and clear the root: $(x^2-1)(y')^2 = m^2 y^2$. Done.

For $y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + \cdots}}}$:

• The expression under the first root is $f(x)$ plus the SAME infinite expression, i.e. $f(x) + y$.
• So $y = \sqrt{f(x) + y}$, giving the finite equation $y^2 = f(x) + y$.
• Differentiate implicitly: $2y\,y' = f'(x) + y'$, so $\dfrac{dy}{dx} = \dfrac{f'(x)}{2y - 1}$.

For $f(x) = x - \sin x$ this gives $\dfrac{dy}{dx} = \dfrac{1 - \cos x}{2y - 1}$.

Two recurring routes:

• Reciprocal substitution. Given $a f(x) + b f(1/x) = g(x)$, replace $x \to 1/x$ to get a second equation, then solve the two simultaneously for $f(x)$.
• Coefficient comparison. Given $f(x) = x^3 + x^2 f'(1) + x f''(2) + 6$, let $f'(1)=a$, $f''(2)=b$ (constants); differentiate to get $f'(x), f''(x)$, evaluate at the stated points, and solve for $a, b$. Here $a=-5, b=2$, so $f(x)=x^3-5x^2+2x+6$ and $f(2)=-2$.

Once $f$ is explicit, differentiate normally.