Choosing a more powerful technique

Guiding question. Can a different description of the same function make its derivative easier to find?

Implicit differentiation: follow a branch

Suppose an equation \(x^2+xy+y^2=7\) determines a differentiable branch \(y=y(x)\) near a point. Differentiate the identity along that branch:

\[2x+y+xy'+2yy'=0, \qquad y'=-\frac{2x+y}{x+2y},\quad x+2y\ne0.\]

At \((1,2)\) the slope is \(-4/5\). The \(y\) in the product \(xy\) is a function of \(x\), so its derivative contributes \(xy'\).

In general, if \(\Phi(x,y(x))=0\) and \(\Phi\) has continuous partial derivatives, the chain rule gives \(\Phi_x+\Phi_y y'=0\). The partial derivatives mean differentiating with respect to one variable while holding the other fixed. If \(\Phi_y\ne0\) at a point of the curve, the implicit function theorem guarantees a differentiable local branch there. We use that guarantee here without proving the theorem.

If \(\Phi_y=0\), the division formula is unavailable. The curve may have a vertical tangent, several branches, or even a perfectly differentiable branch: inspect the relation itself. For \(x^2+y^2=1\) at \((1,0)\), a finite \(y'\) would require \(2+0\cdot y'=0\), which is impossible. For other equations, a zero coefficient need not imply this contradiction.

Inverse differentiation: reverse the relation

Suppose \(f\) has a continuous inverse near \(b=f(a)\), is differentiable at \(a\), and \(f'(a)\ne0\). Put \(y=f(x)\). As \(y\to b\), continuity of the inverse gives \(x\to a\), and

\[\frac{f^{-1}(y)-f^{-1}(b)}{y-b} =\frac{x-a}{f(x)-f(a)} \longrightarrow\frac1{f'(a)}.\]

Thus \((f^{-1})'(b)=1/f'(a)\). The input to the inverse is \(b\), while the derivative of \(f\) is evaluated at \(a=f^{-1}(b)\).

Apply this to the natural logarithm, the inverse of \(e^x\):

\[(\ln x)'=\frac1x,\qquad x>0.\]

For real \(\alpha\) and \(x>0\), define \(x^\alpha=e^{\alpha\ln x}\). The chain rule now proves \((x^\alpha)'=\alpha x^{\alpha-1}\). Also, for a constant \(b>0\), \(b^x=e^{x\ln b}\) has derivative \(b^x\ln b\). Notice the difference between a constant exponent and a constant base.

Logarithmic differentiation: turn products into sums

For a positive differentiable function \(y\), differentiate \(\ln y\):

\[\frac{y'}y=(\ln y)'.\]

For example, let \(y=(1+x^2)^{\sin x}\). Its base is positive for every \(x\), so

\[\ln y=\sin x\ln(1+x^2),\]

and

\[y'=(1+x^2)^{\sin x} \left[\cos x\ln(1+x^2)+\frac{2x\sin x}{1+x^2}\right].\]

In general, for differentiable \(u>0\) and \(v\),

\[\frac d{dx}u(x)^{v(x)} =u(x)^{v(x)}\left[v'(x)\ln u(x)+v(x)\frac{u'(x)}{u(x)}\right].\]

Both the base and the exponent change. Dropping either contribution is a common error. For a nonzero function of either sign, \(\ln\lvert y\rvert\) gives \(y'/y\) locally; zeros still require separate treatment. A variable real power of a negative base is not generally a real-valued function on an interval.

Revisit Lesson 3. Take the logarithm of \(F\) and recover its three-term factored derivative in one line. This supplies a second route to the same answer.

Parametric differentiation: keep the parameter straight

If a curve is given by differentiable functions \(x=X(t)\) and \(y=Y(t)\), and \(X'(t)\ne0\), then on a local branch where \(t\) can be expressed as a differentiable function of \(x\),

\[\frac{dy}{dx}=\frac{Y'(t)}{X'(t)}.\]

For \(X(t)=e^t\cos t\) and \(Y(t)=e^t\sin t\), the slope at \(t=0\) is \(1/1=1\). If \(X'(t)=0\), the quotient cannot be used there; return to the curve or a limit. A second derivative with respect to \(x\) also requires the chain rule:

\[\frac{d^2y}{dx^2}=\frac1{X'(t)}\frac d{dt}\left(\frac{Y'(t)}{X'(t)}\right).\]

Problems: choose the method

4.1 — A tower with three changing copies of \(x\). For \(x>0\), differentiate \(y=x^{(x^x)}\). The parentheses specify the grouping. Explain why applying the constant-exponent power rule is invalid.

Hint

First write \(\ln y=x^x\ln x\). Differentiate \(x^x\) by the same technique.

Solution

Since \((x^x)'=x^x(\ln x+1)\),

\[y'=x^{(x^x)}x^x \left[(\ln x+1)\ln x+\frac1x\right].\]

The exponent \(x^x\) changes with \(x\), so a formula derived for a constant exponent does not apply.

4.2 — A zero denominator with two finite slopes. Consider \(y^2=x^2(x+1)\) near \((0,0)\). Differentiate implicitly away from \(y=0\). Then exhibit two differentiable branches through the origin and find their slopes there. What information did the divided formula fail to provide?

Hint

Use \(y=x\sqrt{1+x}\) and \(y=-x\sqrt{1+x}\) for \(x>-1\) near zero. Avoid replacing \(x\) by \(\lvert x\rvert\) if you want smooth crossing branches.

Solution

Implicit differentiation gives \(2yy'=3x^2+2x\), so \(y'=(3x^2+2x)/(2y)\) when \(y\ne0\). At the origin this becomes the uninformative identity \(0=0\). The displayed branches have derivatives \(1\) and \(-1\) at zero. The equation alone does not select a single branch or slope there.

4.3 — An inverse without a formula. Let \(f(x)=x^3+x\). Show it has a continuous inverse \(g\) on \(\mathbb R\) without solving the cubic. Find \(g'(2)\) and \(g''(2)\).

Hint

For \(u>v\), factor \(f(u)-f(v)\). Then use \(g^3+g=x\) and differentiate twice.

Solution

The difference is \((u-v)(u^2+uv+v^2+1)>0\). The function is continuous, strictly increasing, and tends to opposite infinities at opposite ends, so it is a bijection with a continuous inverse. Since \(f'(x)=3x^2+1>0\), its inverse is differentiable. As \(g(2)=1\),

\[g'=\frac1{3g^2+1},\qquad g''=-\frac{6g}{(3g^2+1)^3}.\]

Consequently \(g'(2)=1/4\) and \(g''(2)=-3/32\). Differentiating the first derivative formula is legitimate because its denominator never vanishes.

4.4 — Two derivatives, two variables. For \(x=t+t^3\) and \(y=t^2\), find \(dy/dx\) and \(d^2y/dx^2\) at \(t=1\). Explain why \(d^2y/dt^2=2\) is not the requested second derivative.

Hint

After differentiating \(2t/(1+3t^2)\) with respect to \(t\), divide once more by \(dx/dt\).

Solution

Since \(X'=1+3t^2>0\),

\[\frac{dy}{dx}=\frac{2t}{1+3t^2},\qquad \frac{d^2y}{dx^2}=\frac{2-6t^2}{(1+3t^2)^3}.\]

At \(t=1\) these are \(1/2\) and \(-1/16\). Changing \(x\) changes \(t\) at the rate \(dt/dx=1/X'\); the second derivative with respect to the parameter omits this conversion.

Checkpoint. Choose a method before calculating: explicit composition, implicit relation, inverse, logarithm, or parameter. Say why its hypotheses hold at the point you need.