NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Parametric differentiation: when \(\displaystyle x\) and \(\displaystyle y\) are both functions of a parameter \(\displaystyle t\), the chain rule gives \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \text{provided } \frac{dx}{dt}\neq 0 . \]Differentiate each equation with respect to \(\displaystyle t\) by the power rule: \[x=2at^{2}\ \Rightarrow\ \frac{dx}{dt}=4at,\qquad y=at^{4}\ \Rightarrow\ \frac{dy}{dt}=4at^{3}. \]Hence \[\frac{dy}{dx}=\frac{4at^{3}}{4at}=t^{2}. \]\(\displaystyle \dfrac{dy}{dx}=t^{2}\), valid for \(\displaystyle a\neq0\) and \(\displaystyle t\neq0\); at \(\displaystyle t=0\) we have \(\displaystyle dx/dt=0\) and the formula does not apply.

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}\), provided \(\displaystyle \dfrac{dx}{d\theta}\neq0\).Differentiate each equation with respect to \(\displaystyle \theta\): \[x=a\cos\theta\ \Rightarrow\ \frac{dx}{d\theta}=-a\sin\theta,\qquad y=b\cos\theta\ \Rightarrow\ \frac{dy}{d\theta}=-b\sin\theta . \]Therefore \[\frac{dy}{dx}=\frac{-b\sin\theta}{-a\sin\theta}=\frac{b}{a}. \]The cancellation of \(\displaystyle \sin\theta\) is legitimate only when \(\displaystyle \sin\theta\neq0\); at \(\displaystyle \theta=n\pi\), \(\displaystyle dx/d\theta=0\).\(\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}\), a constant — which is what one should expect, since the point \(\displaystyle (a\cos\theta,\;b\cos\theta)\) always satisfies \(\displaystyle ay=bx\) and so moves along a straight line.

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\), provided \(\displaystyle \dfrac{dx}{dt}\neq0\).\[x=\sin t\ \Rightarrow\ \frac{dx}{dt}=\cos t . \]For \(\displaystyle y=\cos 2t\) the chain rule (inner function \(\displaystyle 2t\)) gives \[\frac{dy}{dt}=-2\sin 2t . \] Now use \(\displaystyle \sin 2t=2\sin t\cos t\), so that the \(\displaystyle \cos t\) coming from \(\displaystyle dx/dt\) can be cancelled: \[\frac{dy}{dt}=-4\sin t\cos t . \]Therefore \[\frac{dy}{dx}=\frac{-4\sin t\cos t}{\cos t}=-4\sin t . \]\(\displaystyle \dfrac{dy}{dx}=-4\sin t\;(=-4x)\), for \(\displaystyle \cos t\neq0\), i.e. \(\displaystyle t\neq(2n+1)\dfrac{\pi}{2}\).

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\).Write \(\displaystyle y=\dfrac{4}{t}=4t^{-1}\) before differentiating, then apply the power rule: \[\frac{dx}{dt}=4,\qquad \frac{dy}{dt}=4\cdot(-1)t^{-2}=-\frac{4}{t^{2}} . \]Hence \[\frac{dy}{dx}=\frac{-4/t^{2}}{4}=-\frac{1}{t^{2}} . \]Check: here \(\displaystyle xy=4t\cdot\dfrac4t=16\), so \(\displaystyle y=\dfrac{16}{x}\) and \(\displaystyle \dfrac{dy}{dx}=-\dfrac{16}{x^{2}}=-\dfrac{16}{16t^{2}}=-\dfrac{1}{t^{2}}\), as found.\(\displaystyle \dfrac{dy}{dx}=-\dfrac{1}{t^{2}}\), \(\displaystyle t\neq0\).

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}\). Differentiate each coordinate term by term, using the chain rule on the \(\displaystyle 2\theta\) terms (the inner factor \(\displaystyle 2\) is the step most often dropped):\[x=\cos\theta-\cos 2\theta\ \Rightarrow\ \frac{dx}{d\theta}=-\sin\theta+2\sin 2\theta , \] \[y=\sin\theta-\sin 2\theta\ \Rightarrow\ \frac{dy}{d\theta}=\cos\theta-2\cos 2\theta . \]Therefore \[\frac{dy}{dx}=\frac{\cos\theta-2\cos 2\theta}{-\sin\theta+2\sin 2\theta}=\frac{\cos\theta-2\cos 2\theta}{2\sin 2\theta-\sin\theta}. \]\(\displaystyle \dfrac{dy}{dx}=\dfrac{\cos\theta-2\cos 2\theta}{2\sin 2\theta-\sin\theta}\), valid wherever \(\displaystyle 2\sin 2\theta\neq\sin\theta\).

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}\). (This is the cycloid.)\[x=a(\theta-\sin\theta)\ \Rightarrow\ \frac{dx}{d\theta}=a(1-\cos\theta),\qquad y=a(1+\cos\theta)\ \Rightarrow\ \frac{dy}{d\theta}=-a\sin\theta . \]So \[\frac{dy}{dx}=\frac{-a\sin\theta}{a(1-\cos\theta)}=\frac{-\sin\theta}{1-\cos\theta}. \]Simplify with the half-angle identities \(\displaystyle \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\) and \(\displaystyle 1-\cos\theta=2\sin^{2}\frac{\theta}{2}\): \[\frac{dy}{dx}=-\frac{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}{2\sin^{2}\dfrac{\theta}{2}}=-\frac{\cos\dfrac{\theta}{2}}{\sin\dfrac{\theta}{2}}=-\cot\frac{\theta}{2}. \]\(\displaystyle \dfrac{dy}{dx}=-\cot\dfrac{\theta}{2}\), valid for \(\displaystyle \theta\neq 2n\pi\) (there \(\displaystyle 1-\cos\theta=0\), the cusps of the cycloid, where the tangent is vertical).

Write the two coordinates as products, \(\displaystyle x=\sin^{3}t\,(\cos 2t)^{-1/2}\) and \(\displaystyle y=\cos^{3}t\,(\cos 2t)^{-1/2}\), and differentiate by the product and chain rules. First the common factor: \[\frac{d}{dt}(\cos 2t)^{-1/2}=-\tfrac12(\cos 2t)^{-3/2}\cdot(-2\sin 2t)=\frac{\sin 2t}{(\cos 2t)^{3/2}} . \]Then \[\frac{dx}{dt}=\frac{3\sin^{2}t\cos t}{(\cos 2t)^{1/2}}+\sin^{3}t\cdot\frac{\sin 2t}{(\cos 2t)^{3/2}} =\frac{3\sin^{2}t\cos t\cos 2t+\sin^{3}t\sin 2t}{(\cos 2t)^{3/2}} . \] Put \(\displaystyle \sin 2t=2\sin t\cos t\) and take out \(\displaystyle \sin^{2}t\cos t\): \[\frac{dx}{dt}=\frac{\sin^{2}t\cos t\left(3\cos 2t+2\sin^{2}t\right)}{(\cos 2t)^{3/2}} =\frac{\sin^{2}t\cos t\left(1+2\cos 2t\right)}{(\cos 2t)^{3/2}}, \] since \(\displaystyle 2\sin^{2}t=1-\cos 2t\).In the same way \[\frac{dy}{dt}=\frac{-3\cos^{2}t\sin t}{(\cos 2t)^{1/2}}+\cos^{3}t\cdot\frac{\sin 2t}{(\cos 2t)^{3/2}} =\frac{\cos^{2}t\sin t\left(-3\cos 2t+2\cos^{2}t\right)}{(\cos 2t)^{3/2}} =\frac{\cos^{2}t\sin t\left(1-2\cos 2t\right)}{(\cos 2t)^{3/2}}, \] since \(\displaystyle 2\cos^{2}t=1+\cos 2t\).Dividing, the awkward power \(\displaystyle (\cos 2t)^{3/2}\) cancels: \[\frac{dy}{dx}=\frac{\cos^{2}t\sin t\,(1-2\cos 2t)}{\sin^{2}t\cos t\,(1+2\cos 2t)} =\frac{\cos t\,(1-2\cos 2t)}{\sin t\,(1+2\cos 2t)} . \]Now use the triple-angle forms written with \(\displaystyle \cos 2t\): \[\sin 3t=\sin t\,(3-4\sin^{2}t)=\sin t\,(1+2\cos 2t),\qquad \cos 3t=\cos t\,(4\cos^{2}t-3)=\cos t\,(2\cos 2t-1). \] Hence the numerator is \(\displaystyle -\cos 3t\) and the denominator is \(\displaystyle \sin 3t\): \[\frac{dy}{dx}=-\frac{\cos 3t}{\sin 3t}=-\cot 3t . \]\(\displaystyle \dfrac{dy}{dx}=-\cot 3t\). The parametrisation needs \(\displaystyle \cos 2t>0\), which forces \(\displaystyle \cos^{2}t>\tfrac12\) so \(\displaystyle \cos t\neq0\) automatically; the one extra requirement is \(\displaystyle \sin t\neq0\), where \(\displaystyle dx/dt=0\).

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\).The easy one first: \[y=a\sin t\ \Rightarrow\ \frac{dy}{dt}=a\cos t . \]For \(\displaystyle x\) the term to handle carefully is \(\displaystyle \log\tan\frac{t}{2}\); by the chain rule (derivative of \(\displaystyle \log u\) is \(\displaystyle u'/u\), and the inner \(\displaystyle \frac{t}{2}\) contributes \(\displaystyle \frac12\)): \[\frac{d}{dt}\log\tan\frac{t}{2}=\frac{1}{\tan\dfrac{t}{2}}\cdot\sec^{2}\frac{t}{2}\cdot\frac12 =\frac{\cos\dfrac{t}{2}}{\sin\dfrac{t}{2}}\cdot\frac{1}{2\cos^{2}\dfrac{t}{2}} =\frac{1}{2\sin\dfrac{t}{2}\cos\dfrac{t}{2}}=\frac{1}{\sin t}=\mathrm{cosec}\,t . \]Therefore \[\frac{dx}{dt}=a\left(-\sin t+\frac{1}{\sin t}\right)=a\cdot\frac{1-\sin^{2}t}{\sin t}=\frac{a\cos^{2}t}{\sin t}. \]Dividing, \[\frac{dy}{dx}=\frac{a\cos t}{\dfrac{a\cos^{2}t}{\sin t}}=\frac{a\cos t\sin t}{a\cos^{2}t}=\frac{\sin t}{\cos t}=\tan t . \]\(\displaystyle \dfrac{dy}{dx}=\tan t\), valid where \(\displaystyle \tan\dfrac{t}{2}>0\) (so the logarithm exists) and \(\displaystyle \cos t\neq0\) (so \(\displaystyle dx/dt\neq0\)).

Parametric differentiation: \(\displaystyle \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}\). Use the standard derivatives \(\displaystyle \dfrac{d}{d\theta}\sec\theta=\sec\theta\tan\theta\) and \(\displaystyle \dfrac{d}{d\theta}\tan\theta=\sec^{2}\theta\): \[x=a\sec\theta\ \Rightarrow\ \frac{dx}{d\theta}=a\sec\theta\tan\theta,\qquad y=b\tan\theta\ \Rightarrow\ \frac{dy}{d\theta}=b\sec^{2}\theta . \]Hence \[\frac{dy}{dx}=\frac{b\sec^{2}\theta}{a\sec\theta\tan\theta}=\frac{b\sec\theta}{a\tan\theta} =\frac{b}{a}\cdot\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta} =\frac{b}{a}\cdot\frac{1}{\sin\theta}. \]\(\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}\,\mathrm{cosec}\,\theta\), valid for \(\displaystyle \cos\theta\neq0\) (needed for \(\displaystyle \sec\theta,\tan\theta\)) and \(\displaystyle \sin\theta\neq0\) (where \(\displaystyle dx/d\theta=0\)). The curve is the hyperbola \(\displaystyle \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\).

Parametric differentiation, with the product rule on \(\displaystyle \theta\sin\theta\) and \(\displaystyle \theta\cos\theta\).\[\frac{dx}{d\theta}=a\Big(-\sin\theta+\big(1\cdot\sin\theta+\theta\cos\theta\big)\Big)=a\big(-\sin\theta+\sin\theta+\theta\cos\theta\big)=a\theta\cos\theta , \] \[\frac{dy}{d\theta}=a\Big(\cos\theta-\big(1\cdot\cos\theta-\theta\sin\theta\big)\Big)=a\big(\cos\theta-\cos\theta+\theta\sin\theta\big)=a\theta\sin\theta . \] The whole point of the question is this cancellation: the product rule supplies exactly the term that kills the first one, and getting the sign in \(\displaystyle \dfrac{d}{d\theta}(\theta\cos\theta)=\cos\theta-\theta\sin\theta\) right is what makes it happen.Therefore \[\frac{dy}{dx}=\frac{a\theta\sin\theta}{a\theta\cos\theta}=\tan\theta . \]\(\displaystyle \dfrac{dy}{dx}=\tan\theta\), valid for \(\displaystyle \theta\neq0\) and \(\displaystyle \cos\theta\neq0\) (this is the involute of a circle).

Write both coordinates as single powers of \(\displaystyle a\) (take \(\displaystyle a>0\), so the powers are defined): \[x=\sqrt{a^{\sin^{-1}t}}=a^{\frac12\sin^{-1}t},\qquad y=\sqrt{a^{\cos^{-1}t}}=a^{\frac12\cos^{-1}t}. \]Use logarithmic differentiation, since the variable sits in the exponent. Taking natural logarithms, \[\log x=\tfrac12\big(\sin^{-1}t\big)\log a,\qquad \log y=\tfrac12\big(\cos^{-1}t\big)\log a . \]Differentiate each with respect to \(\displaystyle t\), using \(\displaystyle \dfrac{d}{dt}\sin^{-1}t=\dfrac{1}{\sqrt{1-t^{2}}}\) and \(\displaystyle \dfrac{d}{dt}\cos^{-1}t=-\dfrac{1}{\sqrt{1-t^{2}}}\) — that minus sign is the entire content of the result: \[\frac{1}{x}\frac{dx}{dt}=\frac{\log a}{2\sqrt{1-t^{2}}}\ \Rightarrow\ \frac{dx}{dt}=\frac{x\log a}{2\sqrt{1-t^{2}}}, \] \[\frac{1}{y}\frac{dy}{dt}=-\frac{\log a}{2\sqrt{1-t^{2}}}\ \Rightarrow\ \frac{dy}{dt}=-\frac{y\log a}{2\sqrt{1-t^{2}}} . \]Hence, by parametric differentiation, \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt} =\frac{-\dfrac{y\log a}{2\sqrt{1-t^{2}}}}{\dfrac{x\log a}{2\sqrt{1-t^{2}}}} =-\frac{y}{x}, \] which is what was to be shown.Check, independently: \(\displaystyle xy=a^{\frac12(\sin^{-1}t+\cos^{-1}t)}=a^{\frac{\pi}{4}}\), a constant, because \(\displaystyle \sin^{-1}t+\cos^{-1}t=\dfrac{\pi}{2}\). Differentiating \(\displaystyle xy=\) constant implicitly gives \(\displaystyle x\dfrac{dy}{dx}+y=0\), i.e. \(\displaystyle \dfrac{dy}{dx}=-\dfrac{y}{x}\).Valid for \(\displaystyle t\in(-1,1)\) and \(\displaystyle a>0,\ a\neq1\) (if \(\displaystyle a=1\) then \(\displaystyle x=y=1\) and nothing varies).