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Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.4 | Set 3

  • Last Updated : 16 May, 2021
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Question 21. Differentiate (2x2 – 3) sin x with respect to x.

Solution:

We have,

=> y = (2x2 – 3) sin x

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}[(2x^2 - 3) sin x]



On using product rule we get,

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

sinx(4x)+(2x^2-3)cosx

4xsinx+2x^2cosx-3cosx

Question 22. Differentiate x^5(3-6x^{-9}) with respect to x.

Solution:

We have,

=> y = x^5(3-6x^{-9})

On differentiating both sides, we get,



\frac{dy}{dx}=\frac{d}{dx}[x^5(3-6x^{-9})]

On using product rule we get,

(3-6x^{-9})\frac{d}{dx}(x^5)+(x^5)\frac{d}{dx}(3-6x^{-9})

(3-6x^{-9})(5x^4)+(x^5)[-(6)(-9)x^{-10}]

(3-6x^{-9})(5x^4)+(x^5)(54x^{-10})

15x^4-30x^5+54x^{-5}

Question 23. Differentiate x^{-4}(3-4x^{-5}) with respect to x.

Solution:

We have, 

=> y = x^{-4}(3-4x^{-5})

On differentiating both sides, we get,



\frac{dy}{dx}=\frac{d}{dx}[x^{-4}(3-4x^{-5})]

On using product rule we get,

(3-4x^{-5})\frac{d}{dx}(x^{-4})+x^{-4}\frac{d}{dx}(3-4x^{-5})

(3-4x^{-5})(-4x^{-5})+x^{-4}[-(4)(-5)x^{-6}]

(3-4x^{-5})(-4x^{-5})+x^{-4}[20x^{-6}]

-12x^{-5}+16x^{-10}+20x^{-10}

-12x^{-5}+36x^{-10}

-12x^{-5}(1+3x^{-5})

Question 24. Differentiate x^{-3}(5+3x) with respect to x.

Solution:

We have,



=> y = x^{-3}(5+3x)

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}[x^{-3}(5+3x)]

On using product rule we get, 

(5+3x)\frac{d}{dx}(x^{-3})+x^{-3}\frac{d}{dx}(5+3x)

(5+3x)(-3x^{-4})+x^{-3}(3)

-15x^{-4}-9x^{-3}+3x^{-3}

-15x^{-4}-6x^{-3}

-3x^{-3}(5x^{-1}-2)

Question 25. Differentiate \frac{ax+b}{cx+d} with respect to x.

Solution:



We have, 

=> y = \frac{ax+b}{cx+d}

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}[\frac{ax+b}{cx+d}]

On using product rule we get, 

(\frac{1}{cx+d})\frac{d}{dx}(ax+b)+(ax+b)\frac{d}{dx}(\frac{1}{cx+d})

(\frac{1}{cx+d})(a)+(ax+b)\left[\frac{-1}{(cx+d)^{2}}×c\right]

(\frac{a}{cx+d})+(ax+b)\left[\frac{-c}{(cx+d)^{2}}\right]

(\frac{a}{cx+d})-\frac{c(ax+b)}{(cx+d)^{2}}

\frac{a(cx+d)-c(ax+b)}{(cx+d)^2}



\frac{acx+ad-acx-bc}{(cx+d)^2}

\frac{ad-bc}{(cx+d)^2}

Question 26. Differentiate (ax + b)n (cx + d)m with respect to x.

Solution:

We have,  

=> y = (ax + b)n (cx + d)m

On differentiating both sides, we get,

\frac{dy}{dx}=\frac{d}{dx}[(ax + b)^n (cx + d)^m]

On using product rule we get, 

(cx + d)^m\frac{d}{dx}[(ax + b)^n]+(ax + b)^n\frac{d}{dx}[(cx + d)^m]

(cx + d)^m[na(ax + b)^{n-1}]+(ax + b)^n[mc(cx + d)^{m-1}]



na(cx + d)^m(ax + b)^{n-1}+mc(ax + b)^n(cx + d)^{m-1}

(cx + d)^{m-1}(ax + b)^{n-1}[na(cx+d)+mc(ax + b)]

Question 27. Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answer are the same.

Solution:

We have, 

=> y = (1 + 2 tan x) (5 + 4 cos x)

On using product rule we get, 

\frac{d}{dx}[(1 + 2 tan x) (5 + 4 cos x)]=(5 + 4 cos x)\frac{d}{dx}(1+2tanx)+(1+2tanx)\frac{d}{dx}(5+4cosx)

(5+4cosx)(2sec^2x)+(1+2tanx)(-4sinx)

= 10 sec2 x + 8 cos x sec2 x − 4 sin x − 8 sin x tan x

=  10sec^2x+8cosx(\frac{1}{cos^2x})−4sinx−8sinx(\frac{sinx}{cosx})



10sec^2x+\frac{8}{cosx}−4sinx−\frac{8sin^2x}{cosx}

10sec^2x+\frac{8(1-sin^2x)}{cosx}−4sinx

10sec^2x+\frac{8cos^2x}{cosx}−4sinx

= 10 sec2 x + 8 cos x − 4 sin x 

By using an alternate method, we have,

\frac{d}{dx}[(1 + 2 tan x) (5 + 4 cos x)]=\frac{d}{dx}(5+4cosx+10tanx+8tanxcosx)

\frac{d}{dx}(5+4cosx+10tanx+8(\frac{sinx}{cosx})cosx)

\frac{d}{dx}(5+4cosx+10tanx+8sinx)

On using chain rule, we get,

= 0 − 4 sin x + 10 sec2 x + 8 cos x

= 10 sec2 x + 8 cos x − 4 sin x 

Hence proved.

Question 28. Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(i) (3x2 + 2)2

Solution:

We have, 

=> y = (3x2 + 2)2

On using product rule we get,

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

(3x^2+2)(6x)+(3x^2+2)(6x)

= 12x (3x2 + 2)

= 36 x3 + 24x



By using an alternate method, we have,

\frac{d}{dx}[(3x^2+2)^2]=\frac{d}{dx}[9x^4+4+12x^2]

On using chain rule, we get,

= 36 x3 + 0 + 24 x

= 36 x3 + 24x

Hence proved.

(ii) (x + 2)(x + 3)

Solution:

We have,

=> y = (x + 2)(x + 3)

On using product rule we get,

\frac{d}{dx}[(x + 2)(x + 3)]=(x+3)\frac{d}{dx}(x+2)+(x+2)\frac{d}{dx}(x+3)

= (x+3)(1)+(x+2)(1)

= x + 3 + x + 2

= 2x + 5

By using an alternate method, we have,

\frac{d}{dx}[(x + 2)(x + 3)]=\frac{d}{dx}[x^2+5x+6]

On using chain rule, we get,

= 2x + 5

Hence proved.

(iii) (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

Solution:

We have, 

=> y = (3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

On using product rule we get,

\frac{d}{dx}[(3secx−4cosecx)(−2sinx+5cosx)]=(−2sinx+5cosx)\frac{d}{dx}(3secx−4cosecx)+(3secx−4cosecx)\frac{d}{dx}(−2sinx+5cosx)

= (−2 sin x + 5 cos x) (3 sec x tan x + 4 cot x cosec x)+ (3 sec x − 4 cosec x) (−2 cos x − 5 sin x)

= −6 sin x sec x tan x − 8 sin x cot x cosec x + 15 cos x sec x tan x  + 20 cos x cot x cosec x − 6 sec x cos x − 15 sec x sin x + 8 cosec x cos x + 20 cosec x sin x

= −6 tan2 x − 8 cot x + 15 tan x + 20 cot2 x − 6 − 15 tan x + 8 cot x + 20

= − 6 − 6 tan2 x + 20 cot2 x + 20

= −6 (1 + tan2 x) + 20 (cot2 x + 1)

= −6 sec2 x + 20 cosec2 x



By using an alternate method, we have,

\frac{d}{dx}[(3secx−4cosecx)(−2sinx+5cosx)]=\frac{d}{dx}(−6secxsinx+15secxcosx+6cosecxsinx-20cosecxcosx)

\frac{d}{dx}(−6tanx+15+6-20cotx)

\frac{d}{dx}(−6tanx-20cotx+21)

On using chain rule, we get,

= −6 sec2 x − (−20 cosec2 x)

= −6 sec2 x + 20 cosec2 x

Hence proved.

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