Question bank — Implicit differentiation — technique, applications

True — by the Chain Rule, differentiating anything containing $y$ w.r.t. $x$ produces the "toll" $\frac{dy}{dx}$, because $y$'s inside-rate-of-change with respect to $x$ is $\frac{dy}{dx}$, not $1$.

False — $x^2+y^2=25$ defines two branches (upper and lower semicircle); implicit differentiation quietly handles whichever branch your point lies on, since $-x/y$ takes the correct sign from the sign of $y$.

False — because both methods describe the same curve, they must give the same slope at each point; the implicit answer is merely written in mixed $x,y$ form. On the circle's top branch $y=\sqrt{25-x^2}$, substituting this into $-x/y$ gives exactly $-x/\sqrt{25-x^2}$ — the explicit derivative — so the two agree line-for-line.

False — it fails wherever $F_y=0$ (division by zero); that is precisely where the tangent is vertical and the Implicit Function Theorem no longer guarantees $y$ as a function of $x$.

True — both sides are equal functions of $x$, so their derivatives are equal; this is why we're allowed to differentiate both sides at once, constants and all.

False — at $(5,0)$ we'd divide by $y=0$; the tangent is vertical (undefined slope), which the formula signals by blowing up rather than giving a number.

False — Logarithmic Differentiation is implicit differentiation applied to $\ln y=(\text{stuff})$; the $\frac{1}{y}\frac{dy}{dx}$ on the left is just the chain-rule toll on $\ln y$.

False — $y^{\prime \prime }$ is supposed to be a function of the position $(x,y)$ alone, so that you can plug in a point and get a number; but $y^{\prime }$ is not an independent quantity — it is itself the known expression $-x/y$. Leaving $y^{\prime }$ in means your answer secretly still contains an unevaluated derivative, so it isn't yet a usable function of $(x,y)$ — substituting it back is what makes it one.

The chain-rule toll is missing; $y$ is a function of $x$, so $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$, not $2y$. It only looks like $\frac{d}{dx}(x^2)=2x$ because both are squares.

$xy$ is a product of two functions, so the Product Rule applies: $\frac{d}{dx}(xy)=1\cdot y+x\cdot \frac{dy}{dx}=y+x{y}^{\prime }$. Treating $xy$ as "one lump" throws away both terms.

A constant has rate of change zero, so $\frac{d}{dx}(25)=0$; the balance of the equation is preserved precisely because constants vanish.

You cannot solve $\sin (xy)=x+y$ for $y$ in closed form, so there is nothing explicit to differentiate; implicit differentiation is the only available route.

The outer chain rule is only half done — you must multiply by $\frac{d}{dx}(xy)=y+x{y}^{\prime }$, giving $\cos (xy)(y+x{y}^{\prime })$. The inside is a function of $x$, so its derivative isn't $1$.

On the range of $\arcsin$, namely $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$, $\cos y\ge 0$, so only the positive root is correct: $y^{\prime }=\frac{1}{\sqrt{1-x^2}}$, valid on the domain $x\in (-1,1)$ (see Derivatives of Inverse Functions).

The power rule needs a constant exponent; here the exponent is the variable $x$, so the rule doesn't apply. Log first: $\ln y=x\ln x$, giving $y^{\prime }=x^x(\ln x+1)$.

You can plug the point in early, but you must still isolate $y^{\prime }$: $y^{\prime }(27-18)=18-27$ gives $y^{\prime }=-1$; skipping the isolation leaves $y^{\prime }$ undetermined.

Because we differentiate with respect to $x$: $x$ changes at rate $1$ with itself ($\frac{dx}{dx}=1$), while $y$ changes at rate $\frac{dy}{dx}$, an unknown we're solving for.

We never split the curve into $y=+\sqrt{\cdots }$ and $y=-\sqrt{\cdots }$; the single formula $-x/y$ automatically carries the right sign from whichever $y$-value the point has, covering both branches at once (see the branch figure below).

$F_y=0$ means the curve is locally vertical, so $y$ can't be written as a function of $x$ there (Implicit Function Theorem); the division by zero is the algebra warning you of a vertical tangent.

$\ln$ turns exponents into multipliers and products into sums ($\ln (x^x)=x\ln x$), converting a case no elementary rule covers into a product the Product Rule handles.

We derive the formula: setting $\sin y=x$ and differentiating implicitly is the general machine that produces every inverse-function derivative — memorising the result skips understanding why it's $1/\sqrt{1-x^2}$ on $x\in (-1,1)$.

In Related Rates both $x$ and $y$ are functions of time, so we differentiate w.r.t. $t$; each variable then pays a toll ($\frac{dx}{dt}$ or $\frac{dy}{dt}$) — same chain-rule logic, different independent variable.

The normal is perpendicular to the tangent, so its slope is $-1/m$ (see Tangent and Normal Lines); implicit differentiation supplies $m=\frac{dy}{dx}$ even when $y=f(x)$ can't be written.

It blows up (denominator $y\to 0$), which is the correct signal that the slope is undefined/vertical, not a mistake — geometry and algebra agree (see the vertical-tangent figure above).

You'd get $0/0$, an indeterminate form flagging a singular point (like the crossing of the folium of Descartes at the origin), where more than one tangent may exist and the simple formula fails.

$\ln x$ and $x^x$ are only real-valued (in the usual sense) for $x>0$; taking $\ln$ of a non-positive base is undefined, so the logarithmic method is restricted to that domain.

At $y=0$ (the points $(\pm 5,0)$) the tangent is vertical and $y$ isn't locally a function of $x$, so both $y^{\prime }$ and $y^{\prime \prime }$ are undefined there — the $y^3$ in the denominator makes this explicit.

The steps still run formally to $\frac{dy}{dx}=-x/y$, but the equation has no real solutions, so there is no real curve and hence no real tangent — always check the locus is nonempty before trusting a derivative.

The slope is exactly $0$ — a horizontal tangent — which is perfectly valid; only a zero denominator ($F_y=0$) causes trouble, not a zero numerator.