Differentiation (Part-38) | S N De

Differentiation (Part-27) | S N De-Class 12

 

$~15(i)~~\log_ax$

Solution. 

Let $~~y=\log_ax =\log_ex ~\cdot~\log_ae \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(\log_ex ~\cdot~\log_ae)\\=\log_ae~\cdot \frac{d}{dx}(\log_ex) \\=\log_ae \cdot \frac 1x\\=\frac 1x~\log_ae~\text{(ans.)}$

$~(ii)~~\log_xe$

Solution.

 Let $~~y=\log_xe=\frac{1}{\log_ex} \\ \text{or,}~~ y=(\log_ex)^{-1} \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}[(\log_ex)^{-1}]\\=-(\log_ex)^{-1-1}~\cdot \frac{d}{dx}(\log_ex)\\=-\frac{1}{(\log_ex)^2}~\cdot \frac 1x\\=-\frac{1}{x(\log_ex)^2}~~\text{(ans.)}$

$~16(i)~~x^x$

Solution. 

Let $~~~~~~~y=x^x \\ \text{or,}~~ \log_ey=\log_ex^x=x\log_ex\\ \therefore \frac{d}{dx}(\log y)= \frac{d}{dx}(x\log x) \\ \text{or,}~~ \frac 1y~\cdot \frac{dy}{dx}\\=\frac{d}{dx}(x)~\cdot \log x+ x~\cdot \frac{d}{dx}(\log x)\\=1 \cdot \log x+x~\cdot \frac 1x\\=\log x+1 \\ \therefore~\frac{dy}{dx}=y(1+\log x)\\~~~~~~~~~=x^x(1+\log x)~\text{(ans.)}$

$~(ii)~~2e^x-3m^x+10^x$

Solution. 

let $~~y=2e^x-3m^x+10^x \\ \therefore \frac{dy}{dx}\\=2\frac{d}{dx}(e^x)-3~\frac{d}{dx}(m^x)+\frac{d}{dx}(10^x)\\=2~\cdot e^x-3m^x\log_em+10^x\log_e(10)~~\text{(ans.)}$

$~(iii)~~2^{x+2}-e^{x+1}+\log_{10}x$

Solution. 

Let $~~y=2^{x+2}-e^{x+1}+\log_{10}x \\ \text{or,}~~ y=2^{x+2}-e^{x+1}+\log_ex ~\cdot~\log_{10}e \\ \therefore~\frac{dy}{dx}\\=\frac{d}{dx}(2^{x+2})-\frac{d}{dx}(e^{x+1})\\~~~~~+\log_{10}e~\cdot \frac{d}{dx}(\log_ex)\\=2^{x+2}~\cdot \frac{d}{dx}(x+2)\\~~~~~-e^{x+1}~\cdot \frac{d}{dx}(x+1)+\log_{10}e~\cdot \frac 1x\\=2^{x}\cdot 2^2~\cdot(1+0)-e^{x+1}~\cdot(1+0)\\~~~~~+\frac{\log_{10}e}{x} \\=4\cdot2^x-e^{x}\cdot e+\frac{\log_{10}e}{x}~~\text{(ans.)}$

$~17(i)~~x^2+y^2=4$

Solution. 

$~~x^2+y^2=4 \\ \therefore \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(4) \\ \text{or,}~~2x+2y~\frac{dy}{dx}=0 \\ \text{or,}~~ 2\left(x+y~\frac{dy}{dx}\right)=0 \\ \text{or,}~~ x+y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{x}{y}~~\text{(ans.)}$

$~(ii)~~x^2+xy+y^2=100$

Solution. 

$~~x^2+xy+y^2=100 \\ \therefore~\text{Differentiating w.r.t}~~x,\\~~~\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(100) \\ \text{or,}~~2x+\frac{d}{dx}(x)~\cdot y+x~\cdot \frac{dy}{dx}+2y~\frac{dy}{dx}=0 \\ \text{or,}~~ 2x+1 \cdot y+x~\cdot \frac{dy}{dx}+2y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}(x+2y)=-(2x+y)\\ \text{or,}~~ \frac{dy}{dx}=-\frac{2x+y}{x+2y}~~\text{(ans.)} $

To download full solution of DIFFERENTIATION (PART-2) [CLASS-XII] of Exercise-3B, click here 

(iii) $~x^3+y^3=3axy$

Solution. 

$~~x^3+y^3=3axy \\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac{d}{dx}(x^3)+\frac{d}{dx}(y^3)=3a~\frac{d}{dx}(xy) \\ \text{or,}~~3x^2+3y^2~\frac{dy}{dx}=3a~\left[x~\frac{dy}{dx}+y~\frac{d}{dx}(x)\right] \\ \text{or,}~~ 3\left(x^2+y^2~\frac{dy}{dx}\right)=3a \left[x~\frac{dy}{dx}+y~\cdot 1\right] \\ \text{or,}~~ x^2+y^2~\frac{dy}{dx}=ax~\frac{dy}{dx}+ay \\ \text{or,}~~ y^2~\frac{dy}{dx}-ax~\frac{dy}{dx}=ay-x^2 \\ \text{or,}~~ -(ax-y^2)~\frac{dy}{dx}=-(x^2-ay) \\ \text{or,}~~\frac{dy}{dx}=\frac{x^2-ay}{ax-y^2}~~\text{(ans.)}$

(iv)$~x^{2/3}+y^{2/3}=a^{2/3}$

Solution. 

$~~x^{2/3}+y^{2/3}=a^{2/3} \\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac{d}{dx}(x^{2/3})+\frac{d}{dx}(y^{2/3})=\frac{d}{dx}(a^{2/3}) \\ \text{or,}~~\frac 23~x^{\frac 23-1}+\frac 23~y^{\frac 23-1}~\frac{dy}{dx}=0 \\ \text{or,}~~\frac 23\left(x^{-1/3}+y^{-1/3}~\frac{dy}{dx}\right)=0 \\ \text{or,}~~ \frac{1}{x^{1/3}}+\frac{1}{y^{1/3}}~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}} \\ \text{or,}~~ \frac{dy}{dx}=-\left(\frac{y}{x}\right)^{\frac 13}~~\text{(ans.)}$

$~(v)~~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Solution. 

$~~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac{1}{a^2}~\frac{d}{dx}(x^2)+\frac{1}{b^2}~\frac{d}{dx}(y^2)=\frac{d}{dx}(1) \\ \text{or,}~~ \frac{1}{a^2}~\cdot 2x+\frac{1}{b^2}~\cdot 2y~\frac{dy}{dx}=0 \\ \text{or,}~~2\left(\frac{x}{a^2}+\frac{y}{b^2}~\frac{dy}{dx} \right)=0 \\ \text{or,}~~ \frac{y}{b^2}~\frac{dy}{dx}=-\frac{x}{a^2} \\ \text{or,}~~ \frac{dy}{dx}=-\frac{xb^2}{ya^2}~~\text{(ans.)}$

$~(vi)~~x=y\log(xy)$

Solution. 

$~~x=y\log(xy)\\ \text{or,}~~\frac xy=\log(xy) \\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~\frac{d}{dx}(x/y)=\frac{d}{dx}[\log(xy)] \\ \text{or,}~~ \frac{d}{dx}(x)~\cdot \frac 1y+x~\cdot \frac{d}{dx}(y^{-1})=\frac{1}{xy}~\cdot \frac{d}{dx}(xy) \\ \text{or,}~~ 1 \cdot \frac 1y+x~\cdot (-y^{-1-1})~\frac{dy}{dx}\\~~~~=\frac{1}{xy}~\cdot(1 \cdot y+x~\cdot \frac{dy}{dx}) \\ \text{or,}~~ \frac 1y-\frac{x}{y^2}~\frac{dy}{dx}=\frac{y}{xy}+\frac{x}{xy}~\frac{dy}{dx} \\ \text{or,}~~ \frac 1y-\frac{x}{y^2}~\frac{dy}{dx}=\frac 1x+\frac 1y~\frac{dy}{dx} \\ \text{or,}~~ \frac 1y-\frac 1x=\left(\frac{x}{y^2}+\frac 1y\right)~\frac{dy}{dx}\\ \text{or,}~~ \frac{dy}{dx}~\left(\frac{x+y}{y^2}\right)=\frac{x-y}{xy} \\ \text{or,}~~\frac{dy}{dx} =\frac{x-y}{xy}~\cdot ~\frac{y^2}{x+y}\\ \text{or,}~~ \frac{dy}{dx}=\frac{y(x-y)}{x(x+y)}~~\text{(ans.)}$

(vii)  $~ax^2+2hxy+by^2+2gx+2fy+c=0$

Solution. 

$~~ax^2+2hxy+by^2+2gx+2fy+c=0 \\ \therefore ~\text{Differentiating w.r.t}~~x,\\~~a~\frac{d}{dx}(x^2)+2h~\frac{d}{dx}(xy)+b~\frac{d}{dx}(y^2)\\~~~~+2g~\frac{d}{dx}(x)+2f~\frac{dy}{dx}+\frac{d}{dx}(c)=0 \\ \text{or,}~~a~\cdot(2x)+2h~\left(1\cdot y+x~\cdot \frac{dy}{dx}\right)\\~~~~+b~(2y~\frac{dy}{dx})+2g~\cdot 1+2f~\frac{dy}{dx}+0=0\\ \text{or,}~~2ax+2h\left(y+x~\frac{dy}{dx}\right)\\~~~~+2by~\frac{dy}{dx}+2g+2f~\frac{dy}{dx}=0\\ \text{or,}~~2\left[(ax+hy+g)\\~~~~+(hx+by+f)~\frac{dy}{dx}\right]=0 \\ \text{or,}~~(hx+by+f)~\frac{dy}{dx}=-(ax+hy+g) \\ \text{or,}~~\frac{dy}{dx}=-\frac{ax+hy+g}{hx+by+f}~~\text{(ans.)}$

(viii)$~(x^2+y^2)^2=xy$

Solution. 

$~~(x^2+y^2)^2=xy \\ \therefore ~\text{Differentiating w.r.t}~~x, \\~~\frac{d}{dx}(x^2+y^2)^2=\frac{d}{dx}(xy) \\ \text{or,}~~ 2(x^2+y^2)~\cdot \frac{d}{dx}(x^2+y^2)\\~~~=1\cdot y+x \cdot \frac{dy}{dx} \\ \text{or,}~~2(x^2+y^2)~\cdot \left[2x+2y~\frac{dy}{dx}\right]\\~~~=y+x~\frac{dy}{dx} \\ \text{or,}~~4x(x^2+y^2)-y\\~~~=[x-4y(x^2+y^2)]~\frac{dy}{dx} \\ \text{or,}~~-[y-4x(x^2+y^2)]\\~~~=-[4y(x^2+y^2)-x]~\frac{dy}{dx} \\ \text{or,}~~ \frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}~~\text{(ans.)}$