2.4, 3.4 - Implicit differentiation

2.4, 3.4 - Implicit differentiation

An equation can describe a curve without making yy its subject. You will differentiate both sides while treating yy as dependent on xx, collect the resulting derivative terms, and find tangent and normal equations, including horizontal and vertical cases.

Dependent y terms

Can you find a tangent to x2+y2=25x^2+y^2=25 without choosing between y=25−x2y=\sqrt{25-x^2} and y=−25−x2y=-\sqrt{25-x^2}? Differentiate the original equation. The crucial step is to remember that moving along the curve changes yy as xx changes.

Implicit differentiation

Implicit differentiation differentiates an equation relating xx and yy without first making yy the subject. On a differentiable branch, treat yy as a function of xx and use the chain rule on every dependent yy-term.

For example, d(y2)/dx=2y dy/dx\mathrm d(y^2)/\mathrm dx=2y\,\mathrm dy/\mathrm dx, because the outside square contributes 2y2y and its inside function contributes dy/dx\mathrm dy/\mathrm dx. Writing just 2y2y would differentiate with respect to yy, rather than xx.

A dependent power

For a differentiable branch y=y(x)y=y(x),

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

where the real power and its derivative are defined. This is the power rule followed by the chain rule. The power rule is in the formula booklet; this dependent-variable form follows from it.

Differentiate every term, including terms on the right-hand side. A constant differentiates to zero.

Worked example 1: a circle

Question. For x2+y2=25x^2+y^2=25, find dy/dx\mathrm dy/\mathrm dx and the gradient at (3,4)(3,4).

Plan. Differentiate both sides with respect to xx, then isolate the derivative.

2x+2ydydx=0chain rule on y2dydx=−xy(y≠0)\begin{aligned} 2x+2y\frac{\mathrm dy}{\mathrm dx}&=0&&\text{chain rule on }y^2\\ \frac{\mathrm dy}{\mathrm dx}&=-\frac xy&& (y\ne0) \end{aligned}

At (3,4)(3,4), the gradient is −3/4-3/4.

Answer. dy/dx=−x/y\mathrm dy/\mathrm dx=-x/y, where y≠0y\ne0; gradient −3/4-3/4 at (3,4)(3,4).

Check. On the upper branch, y=(25−x2)1/2y=(25-x^2)^{1/2}. Explicit differentiation gives −x/25−x2=−x/y-x/\sqrt{25-x^2}=-x/y, confirming the result.

Change only the coefficient of y2y^2. That coefficient must multiply the whole chain-rule result.

Worked example 2: an ellipse

Question. For x2+4y2=25x^2+4y^2=25, find dy/dx\mathrm dy/\mathrm dx and the gradient at (3,2)(3,2).

Plan. Keep the multiplier 44 when differentiating the dependent square.

2x+8ydydx=0dydx=−x4y(y≠0)\begin{aligned} 2x+8y\frac{\mathrm dy}{\mathrm dx}&=0\\ \frac{\mathrm dy}{\mathrm dx}&=-\frac{x}{4y}&& (y\ne0) \end{aligned}

At (3,2)(3,2), this is −3/(4×2)=−3/8-3/(4\times2)=-3/8.

Answer. dy/dx=−x/(4y)\mathrm dy/\mathrm dx=-x/(4y), where y≠0y\ne0; gradient −3/8-3/8 at (3,2)(3,2).

Check. The point satisfies 9+4(4)=259+4(4)=25. Explicitly, the upper branch is y=1225−x2y=\tfrac12\sqrt{25-x^2}, whose derivative at x=3x=3 is −3/8-3/8.

The same reasoning applies to other outside functions:

ddxln⁡y=1ydydx(y>0),ddxey=eydydx.\frac{\mathrm d}{\mathrm dx}\ln y=\frac1y\frac{\mathrm dy}{\mathrm dx}\quad(y>0), \qquad \frac{\mathrm d}{\mathrm dx}e^y=e^y\frac{\mathrm dy}{\mathrm dx}.

For a radian angle yy, d(sin⁡y)/dx=cos⁡y dy/dx\mathrm d(\sin y)/\mathrm dx=\cos y\,\mathrm dy/\mathrm dx. In each case, keep the derivative of the inside variable.

Watch out

In x2+y2=25x^2+y^2=25, yy is not a constant. The wrong line 2x+2y=02x+2y=0 omits dy/dx\mathrm dy/\mathrm dx and cannot give the changing tangent gradient.

Differentiate another weighted curve.

Collecting the derivative

When a term contains both xx and yy, the product rule is needed as well as the chain rule. For xyxy, both factors change with xx, so

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

The first term differentiates xx; the second differentiates yy.

You may write y′y' for dy/dx\mathrm dy/\mathrm dx while collecting terms. It is the same unknown derivative throughout the calculation. Move every y′y'-term to one side, factor out y′y', and divide by its coefficient when that coefficient is nonzero.

Worked example 3: one mixed product

Question. For x2+y2=xy+7x^2+y^2=xy+7, find dy/dx\mathrm dy/\mathrm dx and the gradient at (1,3)(1,3).

Plan. Use the product rule on xyxy, collect y′y', then substitute the point.

2x+2yy′=y+xy′differentiate both sides(2y−x)y′=y−2xcollect derivative termsy′=y−2x2y−x(2y−x≠0)\begin{aligned} 2x+2yy'&=y+xy'&&\text{differentiate both sides}\\ (2y-x)y'&=y-2x&&\text{collect derivative terms}\\ y'&=\frac{y-2x}{2y-x}&& (2y-x\ne0) \end{aligned}

At (1,3)(1,3), y′=(3−2)/(6−1)=1/5y'=(3-2)/(6-1)=1/5.

Answer. dy/dx=(y−2x)/(2y−x)\mathrm dy/\mathrm dx=(y-2x)/(2y-x), where 2y−x≠02y-x\ne0; gradient 1/51/5 at (1,3)(1,3).

Check. The point satisfies 1+9=3+71+9=3+7. In the differentiated equation, 2+6(1/5)=3+1/52+6(1/5)=3+1/5, so both sides equal 16/516/5.

Square just the xx-factor in the mixed product. The product rule now contributes 2xy2xy instead of yy, and x2y′x^2y' instead of xy′xy'.

Worked example 4: a squared mixed factor

Question. For x2+y2=x2y+7x^2+y^2=x^2y+7, find dy/dx\mathrm dy/\mathrm dx and the gradient at (1,3)(1,3).

Plan. Differentiate x2yx^2y as a product, then collect the derivative terms.

2x+2yy′=2xy+x2y′(2y−x2)y′=2xy−2xy′=2x(y−1)2y−x2(2y−x2≠0)\begin{aligned} 2x+2yy'&=2xy+x^2y'\\ (2y-x^2)y'&=2xy-2x\\ y'&=\frac{2x(y-1)}{2y-x^2}&& (2y-x^2\ne0) \end{aligned}

At (1,3)(1,3), the gradient is 2(1)(2)/(6−1)=4/52(1)(2)/(6-1)=4/5.

Answer. dy/dx=2x(y−1)/(2y−x2)\mathrm dy/\mathrm dx=2x(y-1)/(2y-x^2), where 2y−x2≠02y-x^2\ne0; gradient 4/54/5 at (1,3)(1,3).

Check. The point again satisfies the curve equation. The differentiated equation gives 2+6(4/5)=6+4/52+6(4/5)=6+4/5, with both sides 34/534/5.

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