P3 4.2, P3 4.3 - Quotient and inverse derivatives

P3 4.2, P3 4.3 - Quotient and inverse derivatives

Use the quotient rule from P3 4.2 to differentiate fractions of functions and obtain the derivatives of secant, cosecant and cotangent. Then use P3 4.3 to find a gradient when the curve is given as x in terms of y, taking care over reciprocal rates and inverse branches.

Quotient rule

If both the numerator and denominator are changing, can you divide their derivatives? Try x2/x=xx^2/x=x, for x≠0x\ne0: its derivative is 11, whereas dividing the separate derivatives gives 2x2x. A quotient needs a different rule.

Write y=u/v=uv−1y=u/v=uv^{-1}. Applying the product and chain rules gives

y′=u′v−1−uv−2v′=u′v−uv′v2use a common denominator.\begin{aligned} y'&=u'v^{-1}-uv^{-2}v'\\ &=\frac{u'v-uv'}{v^2} &&\text{use a common denominator}. \end{aligned}

The subtraction comes from differentiating the negative power of vv.

Quotient rule

For differentiable u(x)u(x) and v(x)v(x), with v(x)≠0v(x)\ne0,

ddx(uv)=vu′−uv′v2.\frac{d}{dx}\left(\frac uv\right)=\frac{vu'-uv'}{v^2}.

Multiply the denominator by the numerator's derivative, subtract the numerator times the denominator's derivative, and divide by the original denominator squared. In the formula booklet.

Choose uu and vv before calculating. Keep brackets around each whole contribution to the numerator, especially when v′v' is negative. Simplifying the original function first can sometimes give a shorter route.

Worked example 1: an exponential quotient

Differentiate y=e3x/xy=e^{3x}/x, x≠0x\ne0.

Plan. Use u=e3xu=e^{3x}, v=xv=x, so u′=3e3xu'=3e^{3x} and v′=1v'=1.

dydx=x(3e3x)−e3x(1)x2quotient rule=e3x(3x−1)x2factor the numerator.\begin{aligned} \frac{dy}{dx}&=\frac{x(3e^{3x})-e^{3x}(1)}{x^2} &&\text{quotient rule}\\ &=\frac{e^{3x}(3x-1)}{x^2} &&\text{factor the numerator}. \end{aligned}

Answer. e3x(3x−1)/x2e^{3x}(3x-1)/x^2.

Check. The product form x−1e3xx^{-1}e^{3x} gives −x−2e3x+3x−1e3x-x^{-2}e^{3x}+3x^{-1}e^{3x}, the same derivative after a common denominator is used.

Change only the denominator from xx to x2x^2. Its derivative and its square both change.

Worked example 2: a squared denominator

Differentiate y=e3x/x2y=e^{3x}/x^2, x≠0x\ne0.

Plan. Use v=x2v=x^2, v′=2xv'=2x; the denominator of the quotient rule is now x4x^4.

dydx=x2(3e3x)−e3x(2x)x4=xe3x(3x−2)x4=e3x(3x−2)x3cancel a common nonzero factor.\begin{aligned} \frac{dy}{dx}&=\frac{x^2(3e^{3x})-e^{3x}(2x)}{x^4}\\ &=\frac{xe^{3x}(3x-2)}{x^4} =\frac{e^{3x}(3x-2)}{x^3} &&\text{cancel a common nonzero factor}. \end{aligned}

Answer. e3x(3x−2)/x3e^{3x}(3x-2)/x^3.

Check. At x=1x=1, the derivative is e3≈20.085537e^3\approx20.085537. A central secant of the original quotient with step 0.00010.0001 gives approximately 20.08553720.085537.

Cancellation simplifies a derivative; it does not restore inputs excluded from the original function's domain.

Derivatives of secant cosecant and cotangent

The reciprocal trigonometric functions inherit their derivatives from sine and cosine.

Recall sec⁡x=1/cos⁡x\sec x=1/\cos x, cosec⁡x=1/sin⁡x\operatorname{cosec}x=1/\sin x and cot⁡x=cos⁡x/sin⁡x\cot x=\cos x/\sin x. Angles must be in radians. For secant, the chain rule gives

ddx(cos⁡x)−1=−(cos⁡x)−2(−sin⁡x)=sin⁡xcos⁡2x=sec⁡xtan⁡x.\frac{d}{dx}(\cos x)^{-1} =-(\cos x)^{-2}(-\sin x) =\frac{\sin x}{\cos^2x}=\sec x\tan x.

For cosecant, there is no second negative sign:

ddx(sin⁡x)−1=−cos⁡xsin⁡2x=−cosec⁡xcot⁡x.\frac{d}{dx}(\sin x)^{-1} =-\frac{\cos x}{\sin^2x} =-\operatorname{cosec}x\cot x.

For cotangent, use the quotient rule and sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1:

ddx(cos⁡xsin⁡x)=−sin⁡2x−cos⁡2xsin⁡2x=−cosec⁡2x.\frac{d}{dx}\left(\frac{\cos x}{\sin x}\right) =\frac{-\sin^2x-\cos^2x}{\sin^2x} =-\operatorname{cosec}^2x.

Reciprocal trigonometric derivatives

ddx(sec⁡x)=sec⁡xtan⁡x,ddx(cosec⁡x)=−cosec⁡xcot⁡x,ddx(cot⁡x)=−cosec⁡2x.\begin{aligned} \frac{d}{dx}(\sec x)&=\sec x\tan x,\\ \frac{d}{dx}(\operatorname{cosec}x)&=-\operatorname{cosec}x\cot x,\\ \frac{d}{dx}(\cot x)&=-\operatorname{cosec}^2x. \end{aligned}

Secant requires cos⁡x≠0\cos x\ne0; cosecant and cotangent require sin⁡x≠0\sin x\ne0. In the formula booklet.

For an input kx+ckx+c, keep that input in every factor and multiply the whole derivative by kk. The signs above do not depend on the numerical sign of the function itself.

Worked example 3: a shared linear angle

Differentiate y=sec⁡(2x)+cosec⁡(2x)+cot⁡(2x)y=\sec(2x)+\operatorname{cosec}(2x)+\cot(2x).

Plan. Apply each reciprocal derivative and multiply each by the inner derivative 22.

dydx=2sec⁡(2x)tan⁡(2x)−2cosec⁡(2x)cot⁡(2x)−2cosec⁡2(2x).\frac{dy}{dx}=2\sec(2x)\tan(2x) -2\operatorname{cosec}(2x)\cot(2x) -2\operatorname{cosec}^2(2x).

Answer. The expression above, at inputs where sin⁡(2x)≠0\sin(2x)\ne0 and cos⁡(2x)≠0\cos(2x)\ne0.

Check. Replacing secant, cosecant and cotangent by their sine/cosine fractions gives the same three derivative contributions obtained in the derivations above, each multiplied by 22.

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