Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.5 | Set 1

Last Updated : 30 Apr, 2021

Differentiate the following functions with respect to x:

Question 1. $\frac{x^2+1}{x+1}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{x^2+1}{x+1})=\frac{(x+1)\frac{d}{dx}(x^2+1)-(x^2+1)\frac{d}{dx}(x+1)}{(x+1)^2}$
= $\frac{(x+1)(2x)-x^2-1}{(x+1)^2}$
= $\frac{2x^2+2x-x^2-1}{(x+1)^2}$
= $\frac{x^2+2x-1}{(x+1)^2}$
= $\left(\frac{x-1}{x+1}\right)^2$

Question 2. $\frac{2x-1}{x^2+1}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{2x-1}{x^2+1})=\frac{(x^2+1)\frac{d}{dx}(2x-1)-(2x-1)\frac{d}{dx}(x^2+1)}{(x^2+1)^2}$
= $\frac{2(x^2+1)-2x(2x-1)}{(x^2+1)^2}$
= $\frac{2x^2+2-4x^2+2x}{(x^2+1)^2}$
= $\frac{-2x^2+2x+2}{(x^2+1)^2}$
= $\frac{2(-x^2+x+1)}{(x^2+1)^2}$

Question 3. $\frac{x+e^x}{1+logx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{x+e^x}{1+logx})=\frac{(1+logx)\frac{d}{dx}(x+e^x)-(x+e^x)\frac{d}{dx}(1+logx)}{(1+logx)^2}$
= $\frac{(1+logx)(1+e^x)-(x+e^x)\frac{1}{x}}{(1+logx)^2}$
= $\frac{x(1+logx+e^x+e^xlogx)-x-e^x}{x(1+logx)^2}$
= $\frac{x+xlogx+xe^x+xe^xlogx-x-e^x}{x(1+logx)^2}$
= $\frac{xlogx(1+e^x)-e^x(1-x)}{x(1+logx)^2}$

Question 4. $\frac{e^x-tanx}{cotx-x^n}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{e^x-tanx}{cotx-x^n})=\frac{(cotx-x^n)\frac{d}{dx}(e^x-tanx)-(e^x-tanx)\frac{d}{dx}(cotx-x^n)}{(cotx-x^n)^2}$
= $\frac{(cotx-x^n)(e^x-sec^2x)-(e^x-tanx)(-cosec^2x-nx^{n-1})}{(cotx-x^n)^2}$
= $\frac{(cotx-x^n)(e^x-sec^2x)+(e^x-tanx)(cosec^2x+nx^{n-1})}{(cotx-x^n)^2}$

Question 5. $\frac{ax^2+bx+c}{px^2+qx+r}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{ax^2+bx+c}{px^2+qx+r})=\frac{(px^2+qx+r)\frac{d}{dx}(ax^2+bx+c)-(ax^2+bx+c)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2}$
= $\frac{(px^2+qx+r)(2ax+b)-(ax^2+bx+c)(2px+q)}{(px^2+qx+r)^2}$
= $\frac{2apx^3+2aqx^2+2axr+bpx^2+bqx+br-2apx^3-2bpx^2-2cpx-abx^2-b^2x-bc}{(px^2+qx+r)^2}$
= $\frac{2apx^3+2aqx^2+2axr+bpx^2+bqx+br-2apx^3-2bpx^2-2cpx-aqx^2-bqx-cq}{(px^2+qx+r)^2}$
= $\frac{aqx^2-bpx^2+2arx-2cpx+br-cq}{(px^2+qx+r)^2}$
= $\frac{x^2(aq-bp)+2x(ar-cp)+br-cq}{(px^2+qx+r)^2}$

Question 6. $\frac{x}{1+tanx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{x}{1+tanx})=\frac{(1+tanx)\frac{d}{dx}(x)-(x)\frac{d}{dx}(1+tanx)}{(1+tanx)^2}$
= $\frac{(1+tanx)-x(sec^2x)}{(1+tanx)^2}$
= $\frac{1+tanx-xsec^2x}{(1+tanx)^2}$

Question 7. $\frac{1}{ax^2+bx+c}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{1}{ax^2+bx+c})=\frac{(ax^2+bx+c)\frac{d}{dx}(1)-\frac{d}{dx}(ax^2+bx+c)}{(ax^2+bx+c)^2}$
= $\frac{0-(2ax+b)}{(ax^2+bx+c)^2}$
= $\frac{-2ax-b}{(ax^2+bx+c)^2}$

Question 8. $\frac{e^x}{1+x^2}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{e^x}{1+x^2})=\frac{(1+x^2)\frac{d}{dx}(e^x)-e^x\frac{d}{dx}(1+x^2)}{(1+x^2)^2}$
= $\frac{(1+x^2)e^x-2xe^x}{(1+x^2)^2}$
= $\frac{e^x(1+x^2-2x)}{(1+x^2)^2}$
= $\frac{e^x(1-x)^2}{(1+x^2)^2}$

Question 9. $\frac{e^x+sinx}{1+logx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{e^x+sinx}{1+logx})=\frac{(1+logx)\frac{d}{dx}(e^x+sinx)-(e^x+sinx)\frac{d}{dx}(1+logx)}{(1+logx)^2}$
= $\frac{(1+logx)(e^x+cosx)-(e^x+sinx)\frac{1}{x}}{(1+logx)^2}$
= $\frac{x(1+logx)(e^x+cosx)-(e^x+sinx)}{x(1+logx)^2}$

Question 10. $\frac{xtanx}{secx+tanx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{xtanx}{secx+tanx})=\frac{(secx+tanx)\frac{d}{dx}(xtanx)-(xtanx)\frac{d}{dx}(secx+tanx)}{(secx+tanx)^2}$
= $\frac{(secx+tanx)(xsec^2x+tanx)-(xtanx)(secxtanx+sec^2x)}{(secx+tanx)^2}$
= $\frac{(secx+tanx)(xsec^2x+tanx)-(xtanxsecx)(tanx+secx)}{(secx+tanx)^2}$
= $\frac{(secx+tanx)(xsec^2x+tanx-xtanxsecx)}{(secx+tanx)^2}$
= $\frac{xsec^2x+tanx-xtanxsecx}{secx+tanx}$
= $\frac{xsecx(secx-tanx)+tanx}{secx+tanx}$

Question 11. $\frac{xsinx}{1+cosx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{xsinx}{1+cosx})=\frac{(1+cosx)\frac{d}{dx}(xsinx)-(xsinx)\frac{d}{dx}(1+cosx)}{(1+cosx)^2}$
= $\frac{(1+cosx)(xcos+sinx)-(xsinx)(-sinx)}{(1+cosx)^2}$
= $\frac{(1+cosx)(xcos+sinx)+xsin^2x}{(1+cosx)^2}$
= $\frac{xcosx+sinx+xcos^2x+sinxcosx+xsin^2x}{(1+cosx)^2}$
= $\frac{sinx+sinxcosx+x(sin^2x+cos^2x+cosx)}{(1+cosx)^2}$
= $\frac{sinx(1+cosx)+x(1+cosx)}{(1+cosx)^2}$
= $\frac{(sinx+x)(1+cosx)}{(1+cosx)^2}$
= $\frac{x+sinx}{1+cosx}$

Question 12. $\frac{2^xcotx}{\sqrt{x}}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{2^xcotx}{\sqrt{x}})=\frac{(\sqrt{x})\frac{d}{dx}(2^xcotx)-(2^xcotx)\frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2}$
= $\frac{(\sqrt{x})(-2^xcosec^2x+2^xlog2cotx)-(2^xcotx)(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}$
= $\frac{x(-2^xcosec^2x+2^xlog2cotx)-(2^xcotx)(\frac{1}{2})}{\sqrt{x}(\sqrt{x})^2}$
= $\frac{2^x(-x^xcosec^2x+xlog2cotx-(\frac{1}{2})cotx)}{\sqrt{x}(x)}$
= $\frac{2^x(-x^xcosec^2x+xlog2cotx-(\frac{1}{2})cotx)}{x^\frac{3}{2}}$

Question 13. $\frac{sinx-xcosx}{xsinx+cosx}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{sinx-xcosx}{xsinx+cosx})=\frac{(xsinx+cosx)\frac{d}{dx}(sinx-xcosx)-(sinx-xcosx)\frac{d}{dx}(xsinx+cosx)}{(xsinx+cosx)^2}$
= $\frac{(xsinx+cosx)(cosx-(-xsinx+cosx))-(sinx-xcosx)(xcosx+sinx-sinx)}{(xsinx+cosx)^2}$
= $\frac{(xsinx+cosx)(cosx+xsinx-cosx)-(sinx-xcosx)(xcosx)}{(xsinx+cosx)^2}$
= $\frac{(xsinx+cosx)(xsinx)-(sinx-xcosx)(xcosx)}{(xsinx+cosx)^2}$
= $\frac{x^2sin^2x+xcosxsinx-xsinxcosx+xcos^2x}{(xsinx+cosx)^2}$
= $\frac{x^2(sin^2x+cos^2x)}{(xsinx+cosx)^2}$
= $\frac{x^2}{(xsinx+cosx)^2}$

Question 14. $\frac{x^2-x+1}{x^2+x+1}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{x^2-x+1}{x^2+x+1})=\frac{(x^2+x+1)\frac{d}{dx}(x^2-x+1)-(x^2-x+1)\frac{d}{dx}(x^2+x+1)}{(x^2+x+1)^2}$
= $\frac{(x^2+x+1)(2x-1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}$
= $\frac{2x^3+2x^2+2x-x^2-x-1-(2x^3-2x^2+2x+x^2-x+1)}{(x^2+x+1)^2}$
= $\frac{2x^3+2x^2+2x-x^2-x-1-2x^3+2x^2-2x-x^2+x-1}{(x^2+x+1)^2}$
= $\frac{2x^2-2}{(x^2+x+1)^2}$
= $\frac{2(x^2-1)}{(x^2+x+1)^2}$

Question 15. $\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}$

Solution:

By using quotient rule, we get,
$\frac{d}{dx}(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}})=\frac{(\sqrt{a}-\sqrt{x})\frac{d}{dx}(\sqrt{a}+\sqrt{x})-(\sqrt{a}+\sqrt{x})\frac{d}{dx}(\sqrt{a}-\sqrt{x})}{(\sqrt{a}-\sqrt{x})^2}$
= $\frac{(\sqrt{a}-\sqrt{x})(\frac{1}{2\sqrt{x}})-(\sqrt{a}+\sqrt{x})(\frac{-1}{2\sqrt{x}})}{(\sqrt{a}-\sqrt{x})^2}$
= $\frac{(\sqrt{a}-\sqrt{x})(\frac{1}{2\sqrt{x}})+(\sqrt{a}+\sqrt{x})(\frac{1}{2\sqrt{x}})}{(\sqrt{a}-\sqrt{x})^2}$
= $\frac{(\sqrt{a}-\sqrt{x})+(\sqrt{a}+\sqrt{x})}{(2\sqrt{x})(\sqrt{a}-\sqrt{x})^2}$
= $\frac{\sqrt{a}-\sqrt{x}+\sqrt{a}+\sqrt{x}}{(2\sqrt{x})(\sqrt{a}-\sqrt{x})^2}$
= $\frac{2\sqrt{a}}{(2\sqrt{x})(\sqrt{a}-\sqrt{x})^2}$
= $\frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^2}$