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Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.4 | Set 1

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  • Last Updated : 16 May, 2021
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Question 1. Differentiate x3 sin x with respect to x.

Solution:

We have, 

=> y = x3 sin x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^3sinx)

On using product rule, we get,

= sinx\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(sinx)

= sinx (3x2) + x3 (cosx)

= 3x2 sinx + x3 cosx

= x2 (3 sinx + x cos x)

Question 2. Differentiate x3 ex with respect to x.

Solution:

We have,

=> y = x3 ex

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^3e^x)

On using product rule, we get,

= e^x\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(e^x)

= ex (3x2) + x3 (ex)

= 3x2 ex + x3 ex

= x2 ex (3 + x)

Question 3. Differentiate x2 ex log x with respect to x.

Solution:

We have,

=> y = x2 ex log x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^2e^xlog x)

On using product rule, we get,

= e^xlogx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(e^xlogx)

On using product rule again in the second part of the expression, we get,

= e^xlogx(2x)+x^2[logx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(logx)]

= e^xlogx(2x)+x^2(e^xlogx+\frac{e^x}{x})

= 2xe^xlogx+x^2e^xlogx+xe^x

= xe^x(2logx+xlogx+1)

Question 4. Differentiate xn tan x with respect to x.

Solution:

We have,

=> y = xn tan x 

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^ntanx)

On using product rule, we get,

= tanx\frac{d}{dx}(x^n)+x^n\frac{d}{dx}(tanx)

= tanx(nx^{n-1})+x^n(sec^2x)

= nx^{n-1}tanx+x^nsec^2x

= x^{n-1}(ntanx+xsec^2x)

Question 5. Differentiate xn loga x with respect to x.

Solution:

We have,

=> y = xn loga x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^nlog_ax)

On using product rule, we get,

= log_ax\frac{d}{dx}(x^n)+x^n\frac{d}{dx}(log_ax)

= log_ax(nx^{n-1})+x^n\frac{d}{dx}(\frac{logx}{loga})

= log_ax(nx^{n-1})+\frac{x^n}{loga}(\frac{1}{x})

= nx^{n-1}log_ax+\frac{x^{n-1}}{loga}

= x^{n-1}\left(nlog_ax+\frac{1}{loga}\right)

Question 6. Differentiate (x3+x2+1)sinx with respect to x.

Solution:

We have,

=> y = (x^3+x^2+1)sinx

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}[(x^3+x^2+1)sinx]

On using product rule, we get,

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

= sinx(3x^2+2x)+(x^3+x^2+1)(cosx)

= 3x^2sinx+2xsinx+x^3cosx+x^2cosx+cosx

Question 7. Differentiate sin x cos x with respect to x.

Solution:

We have,

=> y = sin x cos x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(sinxcosx)

On using product rule, we get,

= cosx\frac{d}{dx}(sinx)+(sinx)\frac{d}{dx}(cosx)

= cos x (cos x) − sin x (−sin x)

= cos2 x − sin2 x

= cos2 x − (1 − cos2 x)

= cos2 x − 1 + cos2 x

= 2 cos2 x − 1

= cos 2x

Question 8. Differentiate \frac{2^xcotx}{\sqrt{x}} with respect to x.

Solution:

We have,

=> y = \frac{2^xcotx}{\sqrt{x}}

=> y = 2^x×cotx×x^{\frac{-1}{2}}

On differentiating both sides with respect to x, we get,

\frac{d}{dx}(\frac{2^xcotx}{\sqrt{x}})=\frac{d}{dx}(2^x×cotx×x^{\frac{-1}{2}})

On using product rule, we get,

= \frac{cotx}{\sqrt{x}}\frac{d}{dx}(2^x)+2^x\frac{d}{dx}(cotx×x^{\frac{-1}{2}})

On using product rule again in the second part of the expression, we get,

= \frac{2^xlog2cotx}{\sqrt{x}}+2^x[x^{\frac{-1}{2}}\frac{d}{dx}(cotx)+cotx\frac{d}{dx}(x^{\frac{-1}{2}})]

= \frac{2^xlog2cotx}{\sqrt{x}}+2^x\left[\frac{-cosec^2x}{\sqrt{x}}+\frac{-cotx}{2}\frac{1}{x^{\frac{3}{2}}}\right]

= \frac{2^xlog2cotx}{\sqrt{x}}-\frac{2^xcosec^2x}{\sqrt{x}}-\frac{2^xcotx}{2x^{\frac{3}{2}}}

= \frac{2^x}{\sqrt{x}}\left(log2cotx-cosec^2x-\frac{cotx}{2x}\right)

Question 9. Differentiate x2 sin x log x with respect to x.

Solution:

We have,

=> y = x2 sin x log x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^2sinxlogx)

On using product rule, we get,

= sinxlogx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(sinxlogx)

On using product rule again in the second part of the expression, we get,

= sinxlogx(2x)+x^2[logx\frac{d}{dx}(sinx)+sinx\frac{d}{dx}(logx)]

= 2xsinxlogx+x^2[logxcosx+sinx(\frac{1}{x})]

= 2xsinxlogx+x^2logxcosx+xsinx

Question 10. Differentiate x5 ex + x6 log x with respect to x.

Solution:

We have,

=> y = x5 ex + x6 log x

On differentiating both sides with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}(x^5e^x+x^6log x)

On using chain rule, we get,

= \frac{d}{dx}(x^5e^x)+\frac{d}{dx}(x^6log x)

On using product rule, we get,

= e^x\frac{d}{dx}(x^5)+x^5\frac{d}{dx}(e^x)+logx\frac{d}{dx}(x^6)+x^6\frac{d}{dx}(logx)

= e^x(5x^4)+x^5(e^x)+logx(6x^5)+x^6(\frac{1}{x})

= 5x^4e^x+x^5e^x+6x^5logx+x^5

= x^4(5e^x+xe^x+6xlogx+x)


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