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:

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

Sol:

$\begin{array}{l}Giventhaty={\left(cosx\right)}^{{\left(cosx\right)}^{\left(cosx\right)\dots \infty }}\\ \Rightarrow y={\left(cosx\right)}^y\left[?y={\left(cosx\right)}^{{\left(cosx\right)}^{\left(cosx\right)\dots \infty }}\right]\\ Takinglogonboththesides,logy=y.log\left(cosx\right)\\ Differentiatingbothsidesw.r.t.x\\ \frac{1}{y}.\frac{dy}{dx}=y.\frac{d}{dx}log\left(cosx\right)+log\left(cosx\right).\frac{dy}{dx}\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=y.\frac{1}{cosx}.\frac{d}{dx}\left(cosx\right)+log\left(cosx\right).\frac{dy}{dx}\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=y.\frac{1}{cosx}.\left(-sinx\right)+log\left(cosx\right).\frac{dy}{dx}\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}-log\left(cosx\right).\frac{dy}{dx}=-ytanx\\ \Rightarrow \left[\frac{1}{y}-log\left(cosx\right)\right]\frac{dy}{dx}=-ytanx\\ \Rightarrow \frac{dy}{dx}=\frac{-ytanx}{\frac{1}{y}-log\left(cosx\right)}=\frac{y^{2}tanx}{ylogcosx-1}\\ Hence,\frac{dy}{dx}=\frac{y^{2}tanx}{ylogcosx-1}\\ Hence,proved.\end{array}$

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

Sol:

$\begin{array}{l}Giventhat{y}^x=e^{y-x}\\ Takinglogonboththesides,\\ log{y}^x=log{e}^{y-x}\\ \Rightarrow xlogy=\left(y-x\right)loge\Rightarrow xlogy=y-x\left[?loge=1\right]\\ \Rightarrow xlogy+x=y\\ \Rightarrow x\left(logy+1\right)=y\\ \Rightarrow x=\frac{y}{logy+1}\\ Differentiatingbothsidesw.r.t.y\\ \frac{dx}{dy}=\frac{d}{dy}\left(\frac{y}{logy+1}\right)\\ \Rightarrow =\frac{\left(logy+1\right).1-y.\frac{d}{dy}\left(logy+1\right)}{{\left(logy+1\right)}^2}\\ \Rightarrow =\frac{\left(logy+1\right)-y.\frac{1}{y}}{{\left(logy+1\right)}^2}=\frac{logy+1-1}{{\left(logy+1\right)}^2}=\frac{logy}{{\left(logy+1\right)}^2}\\ Weknowthat\\ Hence,\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\frac{logy}{{\left(logy+1\right)}^2}}=\frac{{\left(logy+1\right)}^2}{logy}\\ Hence,proved.\end{array}$

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

Sol:

$\begin{array}{l}Giventhatx=e^{\frac{x}{y}}\\ Takinglogonboththesides,\\ logx=log{e}^{\frac{x}{y}}\\ \Rightarrow logx=\frac{x}{y}loge\Rightarrow logx=\frac{x}{y}\left[?loge=1\right]\dots \left(1\right)\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}logx=\frac{d}{dx}\left(\frac{x}{y}\right)\\ \Rightarrow \frac{1}{x}=\frac{y.1-x.\frac{dy}{dx}}{y^2}\Rightarrow y^2=x\left(y-x.\frac{dy}{dx}\right)\\ \Rightarrow y^2=xy-x^2.\frac{dy}{dx}\Rightarrow -x^2.\frac{dy}{dx}=y^2-xy\\ \Rightarrow \frac{dy}{dx}=\frac{y^2-xy}{-x^2}=\frac{-y^2+xy}{x^2}=\frac{y\left(x-y\right)}{x^2}\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}.\frac{\left(x-y\right)}{x}=\frac{1}{logx}.\left(\frac{x-y}{x}\right)\left[?logx=\frac{x}{y}fromeqn\left(1\right)\right]\\ Hence,\frac{dy}{dx}=\frac{x-y}{xlogx}.Hence,proved.\end{array}$

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

Sol:

$\begin{array}{l}Giventhata{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(a{x}^2+2hxy+b{y}^2+2gx+2fy+c\right)=\frac{d}{dx}\left(0\right)\\ \Rightarrow a.2x+2h\left(x.\frac{dy}{dx}+y.1\right)+b.2y.\frac{dy}{dx}+2g.1+2f.\frac{dy}{dx}+0=0\\ \Rightarrow 2ax+2hx.\frac{dy}{dx}+2hy+2by.\frac{dy}{dx}+2g+2f.\frac{dy}{dx}=0\\ \Rightarrow 2hx.\frac{dy}{dx}+2by.\frac{dy}{dx}+2f.\frac{dy}{dx}=-2ax-2hy-2g\\ \Rightarrow \left(2hx+2by+2f\right).\frac{dy}{dx}=-2\left(ax+hy+g\right)\\ \Rightarrow \frac{dy}{dx}=\frac{-2\left(ax+hy+g\right)}{2\left(hx+by+f\right)}=\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\\ \Rightarrow \frac{dy}{dx}=\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\\ Now,differentiatingbothsidesw.r.t.y\\ \frac{d}{dy}\left(a{x}^2+2hxy+b{y}^2+2gx+2fy+c\right)=\frac{d}{dy}\left(0\right)\\ \Rightarrow 2ax.\frac{dx}{dy}+2h\left(y.\frac{dx}{dy}+x.1\right)+2by+2g.\frac{dx}{dy}+2f.1+0=0\\ \Rightarrow 2ax.\frac{dx}{dy}+2hy.\frac{dx}{dy}+2hx+2by+2g.\frac{dx}{dy}+2f=0\\ \Rightarrow 2ax\frac{dx}{dy}+2hy.\frac{dx}{dy}+2g.\frac{dx}{dy}=-2hx-2by-2f\\ \Rightarrow \left(2ax+2hy+2g\right).\frac{dx}{dy}=-2\left(hx+by+f\right)\\ \Rightarrow \frac{dx}{dy}=\frac{-2\left(hx+by+f\right)}{2\left(ax+hy+g\right)}=\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\\ \Rightarrow \frac{dx}{dy}=\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\\ \therefore \frac{dy}{dx}.\frac{dx}{dy}=\left[\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\right].\left[\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\right]=1\\ Hence,\frac{dy}{dx}.\frac{dx}{dy}=1.Hence,proved.\end{array}$

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

Sol:

$\begin{array}{l}Giventhat{\left(x^2+y^2\right)}^2=xy\\ \Rightarrow x^4+y^4+2x^2{y}^2=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(y^4\right)+2.\frac{d}{dx}\left(x^2{y}^2\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+2\left[x^2.2y.\frac{dy}{dx}+y^2.2x\right]=x.\frac{dy}{dx}+y.1\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}+4x{y}^2=x.\frac{dy}{dx}+y\\ \Rightarrow 4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}-x.\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \left[4y^3+4x^{2}y-x\right]\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\\ Hence,\frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatta{n}^{-1}\left(x^2+y^2\right)=a\\ \Rightarrow x^2+y^2=tana\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(tana\right)\\ \Rightarrow 2x+2y.\frac{dy}{dx}=0\Rightarrow 2y.\frac{dy}{dx}=-2x\\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}\\ Hence,\frac{dy}{dx}=\frac{-x}{y}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatsec\left(x+y\right)=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sec\left(x+y\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{d}{dx}\left(x+y\right)=x.\frac{dy}{dx}+y.1\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\left(1+\frac{dy}{dx}\right)=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right)+sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}-x.\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \left[sec\left(x+y\right).tan\left(x+y\right)-x\right]\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\\ Hence,\frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\end{array}$

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

Sol:

$\begin{array}{l}Giventhatsinxy+\frac{x}{y}=x^2-y\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sin\left(xy\right)+\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(y\right)\\ \Rightarrow cosxy.\frac{d}{dx}\left(xy\right)+\frac{y.\frac{d}{dx}\left(x\right)-x.\frac{dy}{dx}}{y^2}=2x-\frac{dy}{dx}\\ \Rightarrow cosxy\left[x.\frac{dy}{dx}+y.1\right]+\frac{y.1}{y^2}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}+ycosxy+\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}-\frac{x}{y^2}\frac{dy}{dx}+\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[xcosxy-\frac{x}{y^2}+1\right]\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[\frac{x{y}^{2}cosxy-x+y^2}{y^2}\right]\frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}\\ \Rightarrow \frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}\times \frac{y^2}{x{y}^{2}cosxy-x+y^2}\\ =\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\\ Hence,\frac{dy}{dx}=\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\end{array}$