Ncert Solutions Maths class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=sin{x}^2+si{n}^{2}x+si{n}^2\left(x^2\right)\\ Differentiatingbothsidesw.r.t.x\\ \frac{dy}{dx}=\frac{d}{dx}sin\left(x^2\right)+2sinx.\frac{d}{dx}\left(sinx\right)+2sin\left(x^2\right)\frac{d}{dx}sin\left(x^2\right)\\ =cos{x}^2.\frac{d}{dx}\left(x^2\right)+2sinx.\frac{d}{dx}\left(sinx\right)+2sin\left(x^2\right)\frac{d}{dx}sin\left(x^2\right)\\ =cos{x}^2.2x+2sinx.cosx+2sin{x}^2.cos{x}^2.\frac{d}{dx}\left(x^2\right)\\ =2x.cos{x}^2+sin2x+2sin{x}^2.cos{x}^2.2x\\ =2x.cos{x}^2+sin2x+2xsin2x^2\\ Hence,\frac{dy}{dx}=2x.cos{x}^2+sin2x+2xsin2x^2\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=si{n}^n\left(a{x}^2+bx+c\right)\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{dy}{dx}=\frac{d}{dx}si{n}^n\left(a{x}^2+bx+c\right)\\ \Rightarrow =n.si{n}^{n-1}\left(a{x}^2+bx+c\right).\frac{d}{dx}sin\left(a{x}^2+bx+c\right)\\ =n.si{n}^{n-1}\left(a{x}^2+bx+c\right).cos\left(a{x}^2+bx+c\right).\frac{d}{dx}\left(a{x}^2+bx+c\right)\\ =n.si{n}^{n-1}\left(a{x}^2+bx+c\right).cos\left(a{x}^2+bx+c\right).\left(2ax+b\right)\\ Hence,\frac{dy}{dx}=n\left(2ax+b\right).si{n}^{n-1}\left(a{x}^2+bx+c\right).cos\left(a{x}^2+bx+c\right).\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=log\left[log\left(log{x}^5\right)\right]\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{dy}{dx}=\frac{d}{dx}log\left[log\left(log{x}^5\right)\right]\\ \Rightarrow =\frac{1}{log\left(log{x}^5\right)}\times \frac{d}{dx}log\left(log{x}^5\right)\\ =\frac{1}{log\left(log{x}^5\right)}\times \frac{1}{log\left(x^5\right)}\times \frac{d}{dx}log{x}^5\\ =\frac{1}{log\left(log{x}^5\right)}\times \frac{1}{log\left(x^5\right)}\times \frac{1}{x^5}.\frac{d}{dx}{x}^5\\ =\frac{1}{log\left(log{x}^5\right)}\times \frac{1}{log\left(x^5\right)}\times \frac{1}{x^5}.5x^4\\ =\frac{5}{xlog\left(x^5\right).loglog\left(x^5\right)}\\ Hence,\frac{dy}{dx}=\frac{5}{xlog\left(x^5\right).loglog\left(x^5\right)}.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=\frac{8^x}{x^8}\\ Takinglogonbothsides,weget,logy=log\frac{8^x}{x^8}\\ \Rightarrow logy=log{8}^x-log{x}^8\Rightarrow logy=xlog8-8logx\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log8.1-\frac{8}{x}\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=y\left[log8-\frac{8}{x}\right]\\ Hence,\frac{dy}{dx}=\frac{8^x}{x^8}\left[log8-\frac{8}{x}\right]\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=2^{co{s}^{2}x}\\ Takinglogonbothsides,weget\\ logy=log{2}^{co{s}^{2}x}\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\frac{d}{dx}co{s}^{2}x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\left[2cosx.\frac{d}{dx}cosx\right]\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\left[2cosx\left(-sinx\right)\right]\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2\left(-sin2x\right)\\ \Rightarrow \frac{dy}{dx}=-y.log2sin2x\\ Hence,\frac{dy}{dx}=-2^{co{s}^{2}x}\left(log2sin2x\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf:R\to Rsatisfiestheequationf\left(x+y\right)=f\left(x\right).f\left(y\right)\forall x,y\in R,f\left(x\right)\ne 0.\\ Letustakeanypointx=0atwhichthefunctionf\left(x\right)isdifferentiable.\\ \therefore f\text{'}\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ 2=\underset{h\to 0}{lim}\frac{f\left(0\right).f\left(h\right)-f\left(0\right)}{h}\left[?f\left(0\right)=f\left(h\right)\right]\dots \left(1\right)\\ 2=\underset{h\to 0}{lim}\frac{f\left(0\right).\left[f\left(h\right)-1\right]}{h}\\ Now,f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{lim}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}\left[?f\left(x+y\right)=f\left(x\right).f\left(y\right)\right]\\ =\underset{h\to 0}{lim}\frac{f\left(x\right).\left[f\left(h\right)-1\right]}{h}=2f\left(x\right)\left[Fromeq\left(1\right)\right]\\ Hence,f\text{'}\left(x\right)=2f\left(x\right)\end{array}$