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Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.5 | Set 2

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  • Last Updated : 30 Apr, 2021
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Differentiate the following functions with respect to x:

Question 16. \frac{a+sinx}{1+asinx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{a+sinx}{1+asinx})=\frac{(1+asinx)\frac{d}{dx}(a+sinx)-(a+sinx)\frac{d}{dx}(1+asinx)}{(1+asinx)^2}

= \frac{(1+asinx)(cosx)-(a+sinx)(acosx)}{(1+asinx)^2}

= \frac{cosx+asinxcosx-a^2cosx-asinxcosx}{(1+asinx)^2}

= \frac{cosx-a^2cosx}{(1+asinx)^2}

= \frac{cosx(1-a^2)}{(1+asinx)^2}

Question 17. \frac{10^x}{sinx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{10^x}{sinx})=\frac{(sinx)\frac{d}{dx}(10^x)-(10^x)\frac{d}{dx}(sinx)}{(sinx)^2}

= \frac{(sinx)(10^xlog10)-(10^x)(cosx)}{sin^2x}

= 10x cosec x log 10 − 10x cosec x cot x

= 10x cosec x (log 10 − cot x)

Question 18. \frac{1+3^x}{1-3^x}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{1+3^x}{1-3^x})=\frac{(1-3^x)\frac{d}{dx}(1+3^x)-(1+3^x)\frac{d}{dx}(1-3^x)}{(1-3^x)^2}

= \frac{(1-3^x)(3^xlog3)-(1+3^x)(-3^xlog3)}{(1-3^x)^2}

= \frac{(3^xlog3)(1-3^x+1+3^x)}{(1-3^x)^2}

= \frac{2(3^xlog3)}{(1-3^x)^2}

Question 19. \frac{3^x}{x+tanx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{3^x}{x+tanx})=\frac{(x+tanx)\frac{d}{dx}(3^x)-(3^x)\frac{d}{dx}(x+tanx)}{(x+tanx)^2}

= \frac{(x+tanx)(3^xlog3)-(3^x)(1+sec^2x)}{(x+tanx)^2}

= \frac{3^x[(x+tanx)(log3)-(1+sec^2x)]}{(x+tanx)^2}

Question 20. \frac{1+logx}{1-logx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{1+logx}{1-logx})=\frac{(1-logx)\frac{d}{dx}(1+logx)-(1+logx)\frac{d}{dx}(1-logx)}{(1-logx)^2}

= \frac{(1-logx)\frac{1}{x}-(1+logx)(\frac{-1}{x})}{(1-logx)^2}

= \frac{(1-logx)\frac{1}{x}+(1+logx)(\frac{1}{x})}{(1-logx)^2}

= \frac{1-logx+1+logx}{x(1-logx)^2}

= \frac{2}{x(1-logx)^2}

Question 21. \frac{4x+5sinx}{3x+7cosx}  

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{4x+5sinx}{3x+7cosx})=\frac{(3x+7cosx)\frac{d}{dx}(4x+5sinx)-(4x+5sinx)\frac{d}{dx}(3x+7cosx)}{(3x+7cosx)^2}

= \frac{(3x+7cosx)(4+5cosx)-(4x+5sinx)(3-7sinx)}{(3x+7cosx)^2}

= \frac{12x+15xcosx+28cosx+35cos^2x-(12x-28xsinx+15sinx-35sin^2x)}{(3x+7cosx)^2}

= \frac{12x+15xcosx+28cosx+35cos^2x-12x+28xsinx-15sinx+35sin^2x}{(3x+7cosx)^2}

= \frac{15xcosx+28cosx+28xsinx-15sinx+35(cos^2x+sin^2x)}{(3x+7cosx)^2}

= \frac{15xcosx+28cosx+28xsinx-15sinx+35}{(3x+7cosx)^2}

= \frac{cosx(15x+28)+sinx(28x-15)+35}{(3x+7cosx)^2}

Question 22. \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 23. \frac{a+bsinx}{c+dcosx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{a+bsinx}{c+dcosx})=\frac{(c+dcosx)\frac{d}{dx}(a+bsinx)-(a+bsinx)\frac{d}{dx}(c+dcosx)}{(c+dcosx)^2}

= \frac{(c+dcosx)(bcosx)-(a+bsinx)(-dsinx)}{(c+dcosx)^2}

= \frac{(c+dcosx)(bcosx)+(a+bsinx)(dsinx)}{(c+dcosx)^2}

= \frac{bccosx+bdcos^2x+adsinx+bdsin^2x}{(c+dcosx)^2}

= \frac{bccosx+adsinx+bd(cos^2x+sin^2x)}{(c+dcosx)^2}

= \frac{bccosx+adsinx+bd}{(c+dcosx)^2}

Question 24. \frac{px^2+qx+r}{ax+b}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{px^2+qx+r}{ax+b})=\frac{(ax+b)\frac{d}{dx}(px^2+qx+r)-(px^2+qx+r)\frac{d}{dx}(ax+b)}{(ax+b)^2}

= \frac{(ax+b)(2px+q)-(px^2+qx+r)(a)}{(ax+b)^2}

= \frac{2apx^2+aqx+2bpx+bq-(apx^2+aqx+ar)}{(ax+b)^2}

= \frac{2apx^2+aqx+2bpx+bq-apx^2-aqx-ar}{(ax+b)^2}

= \frac{apx^2+2bpx+bq-ar}{(ax+b)^2}

Question 25. \frac{secx-1}{secx+1}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{secx-1}{secx+1})=\frac{(secx+1)\frac{d}{dx}(secx-1)-(secx-1)\frac{d}{dx}(secx+1)}{(secx+1)^2}

= \frac{(secx+1)(secxtanx)-(secx-1)(secxtanx)}{(secx+1)^2}

= \frac{(secxtanx)(secx+1-secx+1)}{(secx+1)^2}

= \frac{(secxtanx)(2)}{(secx+1)^2}

= \frac{2secxtanx}{(secx+1)^2}

Question 26. \frac{x^5-cosx}{sinx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{x^5-cosx}{sinx})=\frac{(sinx)\frac{d}{dx}(x^5-cosx)-(x^5-cosx)\frac{d}{dx}(sinx)}{(sinx)^2}

= \frac{(sinx)(5x^4+sinx)-(x^5-cosx)(cosx)}{sin^2x}

= \frac{5x^4sinx+sin^2x-(x^5cosx-cos^2x)}{sin^2x}

= \frac{5x^4sinx+sin^2x-x^5cosx+cos^2x}{sin^2x}

= \frac{5x^4sinx-x^5cosx+(sin^2x+cos^2x)}{sin^2x}

= \frac{5x^4sinx-x^5cosx+1}{sin^2x}

Question 27. \frac{x+cosx}{tanx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{x+cosx}{tanx})=\frac{(tanx)\frac{d}{dx}(x+cosx)-(x+cosx)\frac{d}{dx}(tanx)}{(tanx)^2}

= \frac{(tanx)(1+(-sinx))-(x+cosx)(sec^2x)}{tan^2x}

= \frac{(tanx)(1-sinx)-(x+cosx)(sec^2x)}{tan^2x}

Question 28. \frac{x^n}{sinx}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{x^n}{sinx})=\frac{(sinx)\frac{d}{dx}(x^n)-(x^n)\frac{d}{dx}(sinx)}{(sinx)^2}

= \frac{(sinx)(nx^{n-1})-(x^n)(cosx)}{sin^2x}

= \frac{sinx(nx^{n-1})}{sin^2x}-\frac{x^n(cosx)}{sin^2x}

= \frac{nx^{n-1}}{sinx}-\frac{x^n(cosx)}{(sinx)(sinx)}

= nxn-1 cosec x − xn cot x cosec x

Question 29. \frac{ax+b}{px^2+qx+r}

Solution:

By using quotient rule, we get,

\frac{d}{dx}(\frac{ax+b}{px^2+qx+r})=\frac{(px^2+qx+r)\frac{d}{dx}(ax+b)-(ax+b)\frac{d}{dx}(px^2+qx+r)}{(px^2+qx+r)^2}

= \frac{(px^2+qx+r)(a)-(ax+b)(2px+q)}{(px^2+qx+r)^2}

= \frac{apx^2+aqx+ar-(2apx^2+aqx+2bpx+bq)}{(px^2+qx+r)^2}

= \frac{apx^2+aqx+ar-2apx^2-aqx-2bpx-bq}{(px^2+qx+r)^2}

= \frac{-apx^2+ar-2bpx-bq}{(px^2+qx+r)^2}

= \frac{-(apx^2-ar+2bpx+bq)}{(px^2+qx+r)^2}

Question 30. \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{(ax^2+bx+c)(0)-(2ax+b)}{(ax^2+bx+c)^2}

= \frac{-(2ax+b)}{(ax^2+bx+c)^2}


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