Module 11: Implicit, inverse, and logarithmic differentiation

A. Prerequisites and observable goals#

B. The central idea: the dependent variable also changes#

An equation such as $x^2+y^2=25$ describes a relationship, not a single global function $y(x)$. Near a suitable point, however, it may determine one branch of $y$ as a differentiable function of $x$. When differentiating that equation, each appearance of $y$ must be treated as the changing quantity $y(x)$. Thus $(y^2)'=2yy'$, not simply $2y$.

Inverse differentiation asks for the rate of the reverse input-output relationship. Logarithmic differentiation changes a difficult product or variable exponent into a simpler sum or product before applying the derivative. Both methods are applications of established rules, not new kinds of derivative.

C. Conditions that make the methods legitimate#

For a differentiable relation $F(x,y)=0$ on a local differentiable branch, differentiation gives

Here $F_x$ means differentiate the expression in $x$ while temporarily holding $y$ fixed, and $F_y$ means differentiate in $y$ while holding $x$ fixed. This notation is a preview of Module 36; for now you can simply differentiate term by term and collect $y'$. If its denominator is zero, do not conclude a finite slope. Examine the curve and numerator; there may be a vertical tangent or a more complicated singularity.

If $f$ has a differentiable local inverse near $b=f(a)$ and $f'(a)\ne0$, then

A common sufficient setting is a continuously differentiable function near $a$ with nonzero derivative there. The input on the right is $a=f^{-1}(b)$, not $b$. For principal inverse trigonometric functions,

Here $\arcsin$ has range $[-\pi/2,\pi/2]$, $\arccos$ range $[0,\pi]$, and $\arctan$ range $(-\pi/2,\pi/2)$. Inverse notation $\sin^{-1}x$ means $\arcsin x$, not $1/\sin x$. Inverse derivative conditions: [O1Inv].

D. Derivations that expose the branches#

From $f(f^{-1}(x))=x$, the chain rule gives $f'(f^{-1}(x))(f^{-1})'(x)=1$. Division by the nonzero first factor produces the reciprocal rule.

For $y=\arcsin x$, we have $\sin y=x$. Differentiate: $\cos y\,y'=1$. On the interior of the principal range, $\cos y>0$, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$. The positive square root is justified by the branch. Without that range information, the sign would not be determined.

For $y=x^x$ with $x>0$, take logs: $\ln y=x\ln x$. Differentiation gives $y'/y=\ln x+1$, hence $y'=x^x(\ln x+1)$. The usual fixed-power rule alone does not apply because the exponent changes too.

E. How to recognize the right method#

If both $x$ and $y$ appear in an equation and solving for $y$ is awkward, differentiate implicitly. Use product rules for terms such as $xy$ and chain factors for powers of $y$. Collect every $y'$ term before solving. If given data about a function but asked for an inverse derivative, identify which original input produces the requested inverse input. For a variable exponent, first write the function as $e^{v(x)\ln u(x)}$ with $u(x)>0$, or take logs on a positive branch.

F. Worked examples#

Implicit slope. On $x^2+xy+y^2=7$, differentiate to get

Collecting gives $(x+2y)y'=-(2x+y)$. At $(1,2)$, which lies on the curve, $y'=-4/5$. The tangent is $y-2=-\frac45(x-1)$. Checking the point before differentiating prevents computing a “tangent” at a location not on the curve.

Logarithmic differentiation of many factors. For $y=(x^2+1)^3x^{-1/2}$ with $x>0$, write

Then $y'/y=6x/(x^2+1)-1/(2x)$, so multiply back by the original $y$. The derivative of $\ln y$ is $y'/y$, not $1/y$, because $y$ depends on $x$.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 11#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Revisit M03.04 and M08.08. Explain why differentiating $y^2$ with respect to $x$ needs $y'$. Advance when you can use both a given implicit point and inverse-function data without solving an unnecessary cubic equation.