NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

A product of three factors: logarithmic differentiation turns it into a sum.Let \(\displaystyle y=\cos x\cdot\cos 2x\cdot\cos 3x\). Taking logarithms of the modulus of both sides (valid wherever no factor vanishes), \[\log|y|=\log|\cos x|+\log|\cos 2x|+\log|\cos 3x|.\]Differentiate with respect to \(\displaystyle x\), using \(\displaystyle \dfrac{d}{dx}\log|f(x)|=\dfrac{f'(x)}{f(x)}\) and the chain rule on the inner functions \(\displaystyle 2x\) and \(\displaystyle 3x\): \[\frac{1}{y}\frac{dy}{dx}=\frac{-\sin x}{\cos x}+\frac{-2\sin 2x}{\cos 2x}+\frac{-3\sin 3x}{\cos 3x}.\]The inner-derivative factors \(\displaystyle 2\) and \(\displaystyle 3\) are the step most often dropped. Multiplying through by \(\displaystyle y\), \[\frac{dy}{dx}=-\cos x\cdot\cos 2x\cdot\cos 3x\,\bigl(\tan x+2\tan 2x+3\tan 3x\bigr),\] for every \(\displaystyle x\) with \(\displaystyle \cos x\cos 2x\cos 3x\neq 0\).

A quotient inside a square root: take logarithms so the root becomes a factor \(\displaystyle \tfrac12\) and the quotient becomes a difference.Let \[y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}},\] working on a set where the radicand is positive, so \(\displaystyle y>0\). Then \[\log y=\frac{1}{2}\Bigl[\log|x-1|+\log|x-2|-\log|x-3|-\log|x-4|-\log|x-5|\Bigr].\]The three denominator factors carry minus signs; that sign change is the point of the step. Differentiating, \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right].\]Hence \[\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\;\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right],\] valid at every interior point of the domain (in particular \(\displaystyle x\neq 1,2,3,4,5\)).

The base \(\displaystyle \log x\) and the exponent \(\displaystyle \cos x\) both vary, so neither the power rule nor the rule for \(\displaystyle a^{x}\) applies. Take logarithms.Let \(\displaystyle y=(\log x)^{\cos x}\). For the base to be positive we need \(\displaystyle \log x>0\), i.e. \(\displaystyle x>1\). Then \[\log y=\cos x\cdot\log(\log x).\]Differentiate both sides: product rule on the right, and the chain rule on \(\displaystyle \log(\log x)\), whose derivative is \(\displaystyle \dfrac{1}{\log x}\cdot\dfrac{1}{x}\): \[\frac{1}{y}\frac{dy}{dx}=-\sin x\cdot\log(\log x)+\cos x\cdot\frac{1}{x\log x}.\]Therefore \[\frac{dy}{dx}=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\,\log(\log x)\right],\qquad x>1.\]

Differentiate the two terms separately; each has a variable in the exponent, but for different reasons.Let \(\displaystyle y=u-v\) with \(\displaystyle u=x^{x}\) and \(\displaystyle v=2^{\sin x}\), and take \(\displaystyle x>0\) so that \(\displaystyle x^{x}\) is defined.First term (variable base and variable exponent). \(\displaystyle \log u=x\log x\), so by the product rule \[\frac{1}{u}\frac{du}{dx}=\log x+x\cdot\frac{1}{x}=1+\log x,\qquad \frac{du}{dx}=x^{x}(1+\log x).\]Second term (constant base). \(\displaystyle \log v=\sin x\cdot\log 2\), so \[\frac{1}{v}\frac{dv}{dx}=\cos x\,\log 2,\qquad \frac{dv}{dx}=2^{\sin x}\cos x\,\log 2.\] Here \(\displaystyle \log 2\) is a constant and must not be differentiated away; dropping it is the usual error.Hence \[\frac{dy}{dx}=x^{x}(1+\log x)-2^{\sin x}\cos x\,\log 2,\qquad x>0.\]

Logarithmic differentiation converts the triple product into a sum of three simple terms.Let \(\displaystyle y=(x+3)^{2}(x+4)^{3}(x+5)^{4}\). Then \[\log|y|=2\log|x+3|+3\log|x+4|+4\log|x+5|,\] the exponents coming down as multipliers. Differentiating, \[\frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5},\] so \[\frac{dy}{dx}=(x+3)^{2}(x+4)^{3}(x+5)^{4}\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right].\]Cancelling one factor against each denominator gives a polynomial form, which also shows the answer is valid at \(\displaystyle x=-3,-4,-5\) where the logarithms were not: \[\frac{dy}{dx}=(x+3)(x+4)^{2}(x+5)^{3}\Bigl[2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)\Bigr]\] \[=(x+3)(x+4)^{2}(x+5)^{3}\,\bigl(9x^{2}+70x+133\bigr).\]

Two terms, each with a variable exponent; treat them separately. Let \(\displaystyle y=u+v\), where \[u=\left(x+\frac{1}{x}\right)^{x},\qquad v=x^{\left(1+\frac{1}{x}\right)},\qquad x>0.\]First term. \(\displaystyle \log u=x\log\!\left(x+\dfrac{1}{x}\right)\), so by the product rule and chain rule \[\frac{1}{u}\frac{du}{dx}=\log\!\left(x+\frac{1}{x}\right)+x\cdot\frac{1-\dfrac{1}{x^{2}}}{\,x+\dfrac{1}{x}\,}.\] Simplify that fraction before going further — this is where the algebra usually goes wrong: \[\frac{1-\dfrac{1}{x^{2}}}{x+\dfrac{1}{x}}=\frac{(x^{2}-1)/x^{2}}{(x^{2}+1)/x}=\frac{x^{2}-1}{x(x^{2}+1)},\] so the second term is \(\displaystyle x\cdot\dfrac{x^{2}-1}{x(x^{2}+1)}=\dfrac{x^{2}-1}{x^{2}+1}\). Hence \[\frac{du}{dx}=\left(x+\frac{1}{x}\right)^{x}\left[\log\!\left(x+\frac{1}{x}\right)+\frac{x^{2}-1}{x^{2}+1}\right].\]Second term. \(\displaystyle \log v=\left(1+\dfrac{1}{x}\right)\log x\), so \[\frac{1}{v}\frac{dv}{dx}=-\frac{1}{x^{2}}\log x+\left(1+\frac{1}{x}\right)\frac{1}{x}=\frac{x+1-\log x}{x^{2}},\] and therefore \(\displaystyle \dfrac{dv}{dx}=x^{\left(1+\frac{1}{x}\right)}\cdot\dfrac{x+1-\log x}{x^{2}}\).Adding, \[\frac{dy}{dx}=\left(x+\frac{1}{x}\right)^{x}\left[\log\!\left(x+\frac{1}{x}\right)+\frac{x^{2}-1}{x^{2}+1}\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right],\qquad x>0.\]

Let \(\displaystyle y=u+v\) with \(\displaystyle u=(\log x)^{x}\) and \(\displaystyle v=x^{\log x}\). The first needs \(\displaystyle \log x>0\), so take \(\displaystyle x>1\).First term. \(\displaystyle \log u=x\log(\log x)\). Differentiating by the product rule, with \(\displaystyle \dfrac{d}{dx}\log(\log x)=\dfrac{1}{x\log x}\), \[\frac{1}{u}\frac{du}{dx}=\log(\log x)+x\cdot\frac{1}{x\log x}=\log(\log x)+\frac{1}{\log x},\] so \[\frac{du}{dx}=(\log x)^{x}\left[\log(\log x)+\frac{1}{\log x}\right].\]Second term. \(\displaystyle \log v=\log x\cdot\log x=(\log x)^{2}\). Note the exponent is itself \(\displaystyle \log x\), so the logarithm of \(\displaystyle v\) is a square, not a product of two different things: \[\frac{1}{v}\frac{dv}{dx}=2\log x\cdot\frac{1}{x},\qquad \frac{dv}{dx}=x^{\log x}\cdot\frac{2\log x}{x}=2x^{\log x-1}\log x.\]Hence \[\frac{dy}{dx}=(\log x)^{x}\left[\log(\log x)+\frac{1}{\log x}\right]+2x^{\log x-1}\log x,\qquad x>1.\]

Let \(\displaystyle y=u+v\) with \(\displaystyle u=(\sin x)^{x}\) and \(\displaystyle v=\sin^{-1}\sqrt{x}\). The base \(\displaystyle \sin x\) must be positive and \(\displaystyle \sqrt{x}\) must lie in \(\displaystyle [-1,1]\), so work on \(\displaystyle 0<x<1\) (where indeed \(\displaystyle \sin x>0\)).First term, by logarithmic differentiation. \(\displaystyle \log u=x\log\sin x\), so \[\frac{1}{u}\frac{du}{dx}=\log\sin x+x\cdot\frac{\cos x}{\sin x}=\log\sin x+x\cot x,\] \[\frac{du}{dx}=(\sin x)^{x}\bigl(x\cot x+\log\sin x\bigr).\]Second term, by the chain rule with \(\displaystyle \dfrac{d}{dt}\sin^{-1}t=\dfrac{1}{\sqrt{1-t^{2}}}\) and \(\displaystyle t=\sqrt{x}\), \(\displaystyle \dfrac{dt}{dx}=\dfrac{1}{2\sqrt{x}}\): \[\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}\sqrt{1-x}}=\frac{1}{2\sqrt{x-x^{2}}}.\] Squaring \(\displaystyle \sqrt{x}\) to \(\displaystyle x\) inside the root is the step to get right.Therefore \[\frac{dy}{dx}=(\sin x)^{x}\bigl(x\cot x+\log\sin x\bigr)+\frac{1}{2\sqrt{x-x^{2}}},\qquad 0<x<1.\]

Both terms have a variable base and a variable exponent, so each is handled by taking logarithms separately. Let \(\displaystyle y=u+v\) with \(\displaystyle u=x^{\sin x}\), \(\displaystyle v=(\sin x)^{\cos x}\), on a set where \(\displaystyle x>0\) and \(\displaystyle \sin x>0\).First term. \(\displaystyle \log u=\sin x\cdot\log x\), so \[\frac{1}{u}\frac{du}{dx}=\cos x\log x+\sin x\cdot\frac{1}{x},\qquad \frac{du}{dx}=x^{\sin x}\left(\cos x\,\log x+\frac{\sin x}{x}\right).\]Second term. \(\displaystyle \log v=\cos x\cdot\log\sin x\), so \[\frac{1}{v}\frac{dv}{dx}=-\sin x\,\log\sin x+\cos x\cdot\frac{\cos x}{\sin x}=\cos x\cot x-\sin x\,\log\sin x,\] \[\frac{dv}{dx}=(\sin x)^{\cos x}\bigl(\cos x\cot x-\sin x\,\log\sin x\bigr).\]Hence \[\frac{dy}{dx}=x^{\sin x}\left(\frac{\sin x}{x}+\cos x\,\log x\right)+(\sin x)^{\cos x}\bigl(\cos x\cot x-\sin x\,\log\sin x\bigr).\]

Let \(\displaystyle y=u+v\) with \(\displaystyle u=x^{x\cos x}\) and \(\displaystyle v=\dfrac{x^{2}+1}{x^{2}-1}\).First term, by logarithmic differentiation (\(\displaystyle x>0\)). \(\displaystyle \log u=x\cos x\cdot\log x\), a product of three functions, so \[\frac{1}{u}\frac{du}{dx}=(1)\cos x\log x+x(-\sin x)\log x+x\cos x\cdot\frac{1}{x}\] \[=\cos x\log x-x\sin x\log x+\cos x.\] Hence \[\frac{du}{dx}=x^{x\cos x}\bigl[\cos x\,(1+\log x)-x\sin x\,\log x\bigr].\]Second term, by the quotient rule: \[\frac{dv}{dx}=\frac{(2x)(x^{2}-1)-(x^{2}+1)(2x)}{(x^{2}-1)^{2}}=\frac{2x\bigl[(x^{2}-1)-(x^{2}+1)\bigr]}{(x^{2}-1)^{2}}=\frac{-4x}{(x^{2}-1)^{2}}.\] The two \(\displaystyle x^{2}\) terms cancel; keeping the minus sign on the second bracket is essential.Therefore \[\frac{dy}{dx}=x^{x\cos x}\bigl[\cos x\,(1+\log x)-x\sin x\,\log x\bigr]-\frac{4x}{(x^{2}-1)^{2}},\qquad x>0,\ x\neq 1.\]