P4 5.1 - Implicit and parametric differentiation

P4 5.1 - Implicit and parametric differentiation

Find gradients without first writing a curve as y=f(x)y=f(x). Extend the chain rule to implicit and parametric descriptions, use the gradients for tangents and normals, and investigate points where a derivative formula cannot be used directly.

Implicit derivatives

How can you find a circle's tangent without choosing a square-root branch first? In an equation such as x2+y2=7x^2+y^2=7, treat yy locally as a function of xx and differentiate the relationship itself.

Implicit differentiation

Implicit differentiation differentiates an equation linking xx and yy, keeping the dependence of yy on xx. The chain rule gives

ddx(yn)=nyn−1dydx.\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}.

A factor dy/dxdy/dx is needed because the inside variable is yy, rather than xx.

The same principle gives d(ey)/dx=ey(dy/dx)d(e^y)/dx=e^y(dy/dx), d(sin⁡y)/dx=cos⁡y(dy/dx)d(\sin y)/dx=\cos y(dy/dx), and d(ln⁡y)/dx=(1/y)(dy/dx)d(\ln y)/dx=(1/y)(dy/dx) where y>0y>0. A product such as xyxy also needs the product rule:

ddx(xy)=y+xdydx.\frac{d}{dx}(xy)=y+x\frac{dy}{dx}.

Collect all derivative terms before dividing to make dy/dxdy/dx the subject.

Worked example 1: a circle

Question. Find dy/dxdy/dx for x2+y2=7x^2+y^2=7.

Plan. Differentiate both sides with respect to xx, using the chain rule on y2y^2.

2x+2ydydx=0,2ydydx=−2x.2x+2y\frac{dy}{dx}=0, \qquad 2y\frac{dy}{dx}=-2x.

Answer. dy/dx=−x/ydy/dx=-x/y, wherever y≠0y\ne0.

Check. On the upper branch y=7−x2y=\sqrt{7-x^2}, direct chain-rule differentiation gives −x/7−x2=−x/y-x/\sqrt{7-x^2}=-x/y, confirming the result.

Add only a mixed term xyxy to the relationship. Both xx and yy change, so that term contributes twice.

Worked example 2: a mixed term

Question. Find dy/dxdy/dx for x2+xy+y2=7x^2+xy+y^2=7, then find its value at (1,2)(1, 2).

Plan. Apply the product rule to xyxy, then factor the terms containing dy/dxdy/dx.

2x+y+xdydx+2ydydx=0(x+2y)dydx=−2x−ydydx=−2x+yx+2y,x+2y≠0.\begin{aligned} 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}&=0\\ (x+2y)\frac{dy}{dx}&=-2x-y\\ \frac{dy}{dx}&=-\frac{2x+y}{x+2y}, \qquad x+2y\ne0. \end{aligned}

At (1,2)(1, 2), the gradient is −(2+2)/(1+4)=−4/5-(2+2)/(1+4)=-4/5.

Answer. dy/dx=−(2x+y)/(x+2y)dy/dx=-(2x+y)/(x+2y); at (1,2)(1, 2), it is −4/5-4/5.

Check. The point satisfies 1+2+4=71+2+4=7. Locally the upper branch is y=(−x+28−3x2)/2y=(-x+\sqrt{28-3x^2})/2; differentiating it at x=1x=1 gives −1/2−3/10=−4/5-1/2-3/10=-4/5.

A minus sign outside a product applies to both product-rule terms.

Watch out

d(−6xy)/dx=−6y−6x(dy/dx)d(-6xy)/dx=-6y-6x(dy/dx). Writing −6x(dy/dx)+6y-6x(dy/dx)+6y changes a sign incorrectly; writing only −6x(dy/dx)-6x(dy/dx) omits a whole contribution.

Keep the derivative terms together so the requested unsimplified ratio remains visible.

Parametric derivatives

For a parametric curve, compare how fast the two coordinates change with the same parameter. Dividing the vertical rate by the horizontal rate gives the gradient.

The chain rule says dy/dt=(dy/dx)(dx/dt)dy/dt=(dy/dx)(dx/dt). Rearranging requires dx/dt≠0dx/dt\ne0; points violating that condition need separate attention later in the lesson.

Parametric derivative

If x=f(t)x=f(t), y=g(t)y=g(t), and dx/dt≠0dx/dt\ne0,

dydx=dy/dtdx/dt.\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.

Differentiate each coordinate with respect to tt, then divide in this order.

Not in the formula booklet: learn it.

The variable being differentiated matters: dy/dtdy/dt alone is a vertical rate, not usually the gradient of the curve.

Worked example 3: polynomial coordinates

Question. Find dy/dxdy/dx for x=t2+1,y=t3x=t^2+1, y=t^3.

Plan. Differentiate both coordinates and divide vertical rate by horizontal rate.

dxdt=2t,dydt=3t2,dydx=3t22t=32t(t≠0).\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2, \qquad \frac{dy}{dx}=\frac{3t^2}{2t}=\frac32t\quad(t\ne0).

Answer. dy/dx=3t/2dy/dx=3t/2, obtained directly for t≠0t\ne0.

Check. On the positive-tt branch, y=(x−1)3/2y=(x-1)^{3/2}, whose derivative is 3x−1/2=3t/23\sqrt{x-1}/2=3t/2. The negative branch gives the same parameter formula after its sign is included.

Add only a linear term to the vertical coordinate. Its derivative must be added before dividing.

Worked example 4: another contribution to the vertical rate

Question. Find dy/dxdy/dx for x=t2+1,y=t3+tx=t^2+1, y=t^3+t.

Plan. Keep dx/dt=2tdx/dt=2t and differentiate both vertical terms.

dydt=3t2+1,dydx=3t2+12t(t≠0).\frac{dy}{dt}=3t^2+1, \qquad\frac{dy}{dx}=\frac{3t^2+1}{2t}\quad(t\ne0).

Answer. (3t2+1)/(2t)(3t^2+1)/(2t), for t≠0t\ne0.

Check. At t=1t=1, the gradient is 22. The Cartesian positive branch y=(x−1)3/2+(x−1)1/2y=(x-1)^{3/2}+(x-1)^{1/2} has derivative 3/2+1/2=23/2+1/2=2 at x=2x=2.

More complicated coordinates still use your existing differentiation rules. For example, x=t/(1+t)=1−(1+t)−1x=t/(1+t)=1-(1+t)^{-1} has dx/dt=(1+t)−2dx/dt=(1+t)^{-2}, for t≠−1t\ne-1. Differentiate a rational coordinate carefully before taking the rate ratio.

Watch out

The gradient is (dy/dt)/(dx/dt)(dy/dt)/(dx/dt), not the reciprocal. Keep both separate derivatives visible before simplifying.

Use both coordinate rates before substituting the parameter value.

Tangents and normals

Once a gradient and a point are known, the tangent and normal equations use the straight-line methods you already know. Calculate the coordinates as well as the gradient; a parameter value is not a coordinate pair.

For a finite nonzero tangent gradient mm, the tangent through (a,b)(a, b) is y−b=m(x−a)y-b=m(x-a). The normal gradient is −1/m-1/m. Horizontal and vertical tangents need their own line forms, considered in the next part.

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