2.3 - Second derivatives in other forms

2.3 - Second derivatives in other forms

A second derivative measures how a curve's gradient changes with its horizontal coordinate. You will obtain it from implicit and parametric equations, then use its sign at points where the curve can be treated locally as a function of xx.

Implicit second derivatives

Can two points with horizontal tangents on the same implicit curve be a maximum and a minimum? To decide, we need a second derivative without first making yy the subject.

Recall that yy depends on xx. On x2+y2=25x^2+y^2=25, the first differentiation gives

2x+2yy′=0,y′=−xy(y≠0).2x+2yy'=0, \qquad y'=-\frac{x}{y}\quad(y\ne0).

Differentiate the unsimplified equation again. The derivative of yy′yy' needs the product rule:

ddx(yy′)=(y′)2+yy′′.\frac{\mathrm d}{\mathrm dx}(yy')=(y')^2+yy''.

One factor changes at rate y′y', and the other changes at rate y′′y''. Consequently,

2+2(y′)2+2yy′′=0.2+2(y')^2+2yy''=0.

This equation can be evaluated at a point after the first derivative there has been found.

Implicit second differentiation

Treat yy, y′y' and higher derivatives as functions of xx. In particular,

ddx(y2)=2yy′,ddx(yy′)=(y′)2+yy′′.\frac{\mathrm d}{\mathrm dx}(y^2)=2yy', \qquad \frac{\mathrm d}{\mathrm dx}(yy')=(y')^2+yy''.

Differentiate every term in the first-derivative equation, then solve for y′′y'' where its coefficient is nonzero.

These implicit second-derivative steps are not listed as a formula in the formula booklet; they use the supplied product rule and ordinary chain rule.

Keeping the first differentiated equation often makes the second step shorter than applying a quotient rule to a fraction containing yy.

Worked example 1: a circle

Question. Find d2y/dx2\mathrm d^2y/\mathrm dx^2 at (3,4)(3,4) on x2+y2=25x^2+y^2=25.

Plan. Find the first derivative at the point, differentiate the implicit equation again, then substitute.

The point is on the circle because 32+42=253^2+4^2=25. From 2x+2yy′=02x+2yy'=0,

y′=−34.y'=-\frac34.

Differentiate once more and substitute:

2+2(y′)2+2yy′′=01+916+4y′′=0y′′=−2564.\begin{aligned} 2+2(y')^2+2yy''&=0\\ 1+\frac9{16}+4y''&=0\\ y''&=-\frac{25}{64}. \end{aligned}

The point has y≠0y\ne0, so solving for a finite derivative is valid.

Answer. d2ydx2=−2564\displaystyle \frac{\mathrm d^2y}{\mathrm dx^2}=-\frac{25}{64} at (3,4)(3,4).

Check. Near this point, the upper branch is y=25−x2y=\sqrt{25-x^2}. Direct differentiation of this explicit function gives y′′=−25/(25−x2)3/2y''=-25/(25-x^2)^{3/2}, which is −25/64-25/64 at x=3x=3.

Change the coefficient of y2y^2 from 11 to 44. The product-rule structure stays the same, but its coefficient must survive both differentiations.

Worked example 2: an ellipse

Question. Find d2y/dx2\mathrm d^2y/\mathrm dx^2 at (3,2)(3,2) on x2+4y2=25x^2+4y^2=25.

Plan. Differentiate the full implicit equation twice, retaining the factor 44.

The first derivative equation is

2x+8yy′=0,y′=−x4y=−382x+8yy'=0, \qquad y'=-\frac{x}{4y}=-\frac38

at the given point. Differentiating this equation gives

2+8(y′)2+8yy′′=0.2+8(y')^2+8yy''=0.

At (3,2)(3,2),

2+8(964)+16y′′=0⟹y′′=−25128.2+8\left(\frac9{64}\right)+16y''=0 \quad\Longrightarrow\quad y''=-\frac{25}{128}.

Answer. d2ydx2=−25128\displaystyle \frac{\mathrm d^2y}{\mathrm dx^2}=-\frac{25}{128}.

Check. The explicit upper branch is y=1225−x2y=\tfrac12\sqrt{25-x^2}. Its second derivative is half the circle's, so at x=3x=3 it is −25/128-25/128. This also checks the changed coefficient geometrically: the ellipse is vertically compressed by a factor of 1/21/2.

Dividing by yy was justified at these points. At a point with y=0y=0, the circle has a vertical tangent, so the same expression cannot provide a finite dy/dx\mathrm dy/\mathrm dx.

Watch out

Writing (yy′)′=yy′′(yy')'=yy'' loses the term (y′)2(y')^2. Both factors depend on xx; use the product rule before substituting a point.

Retain the new coefficient through both differentiations.

Parametric second derivatives

A parametric derivative first measures change with respect to the parameter. Converting that rate into change with respect to xx requires a division each time.

For example, x=t2x=t^2, y=t3y=t^3 gives

dydx=3t22t=32t(t≠0).\frac{\mathrm dy}{\mathrm dx} =\frac{3t^2}{2t}=\frac32t \qquad(t\ne0).

Differentiating this gradient with respect to tt gives 3/23/2. But that is the gradient's rate of change per unit of tt, not per unit of xx. Divide by dx/dt=2t\mathrm dx/\mathrm dt=2t to get the second derivative 3/(4t)3/(4t).

Parametric second derivative

If x=x(t)x=x(t), y=y(t)y=y(t) are twice differentiable and dx/dt≠0\mathrm dx/\mathrm dt\ne0, then

dydx=dy/dtdx/dt,d2ydx2=ddt(dydx)dx/dt.\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}, \qquad \frac{\mathrm d^2y}{\mathrm dx^2} =\frac{\displaystyle\frac{\mathrm d}{\mathrm dt} \left(\frac{\mathrm dy}{\mathrm dx}\right)} {\mathrm dx/\mathrm dt}.

The first-derivative ratio is in the formula booklet. The second-derivative step is not listed there: apply the same chain-rule conversion again.

Do not replace the numerator by d2y/dt2\mathrm d^2y/\mathrm dt^2: the object to differentiate is the whole first derivative dy/dx\mathrm dy/\mathrm dx.

Worked example 3: differentiating a gradient

Question. A curve is defined by x=t2x=t^2, y=t3y=t^3. Find dy/dx\mathrm dy/\mathrm dx and d2y/dx2\mathrm d^2y/\mathrm dx^2 at t=2t=2.

Plan. Form the first-derivative ratio as a function of tt, differentiate it, then divide by dx/dt\mathrm dx/\mathrm dt.

For t≠0t\ne0,

dydx=3t22t=32t.\frac{\mathrm dy}{\mathrm dx} =\frac{3t^2}{2t}=\frac32t.

Therefore

d2ydx2=3/22t=34t.\frac{\mathrm d^2y}{\mathrm dx^2} =\frac{3/2}{2t}=\frac3{4t}.

At t=2t=2, dx/dt=4≠0\mathrm dx/\mathrm dt=4\ne0, so both calculations apply.

Answer. dy/dx=3\mathrm dy/\mathrm dx=3 and d2y/dx2=3/8\mathrm d^2y/\mathrm dx^2=3/8.

Check. Near positive t=2t=2, eliminate the parameter to get y=x3/2y=x^{3/2}. Directly, y′=3x/2y'=3\sqrt{x}/2 and y′′=3/(4x)y''=3/(4\sqrt{x}). At x=4x=4, these give 33 and 3/83/8.

Stretch only the horizontal coordinate by a factor of 22. The first derivative halves, while the second derivative is divided by 44: both differentiations are with respect to the stretched coordinate.

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