# Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.3

Last Updated : 09 Mar, 2022

### Question 1. Differentiate f(x) = x4 – 2sinx + 3cosx with respect to x.

Solution:

Given that, f(x) = x4 – 2sinx + 3cosx

Now, differentiate w.r.t. x, we get

â‡’ d(x4 – 2sinx + 3cosx) / dx

â‡’ d(x4)/dx – 2.d(sinx)/dx + 3.d(cosx)/dx

â‡’ 4x3 – 2cosx – 3sinx.

### Question 2. Differentiate f(x) = 3x + x3 + 33 with respect to x.

Solution:

Given that, f(x) = 3x + x3 + 33

Now, differentiate w.r.t. x, we get

â‡’ d(3x + x3 + 33) / dx

â‡’ d(3x)/dx + d(x3)/dx + d(33)/dx

â‡’ 3x log3 + 3x2 + 0    [As we know, d(ax)/dx = ax loga]

â‡’ 3x log3 + 3x2

### Question 3. Differentiate f(x) = x3/3 – 2âˆšx + 5/x2 with respect to x.

Solution:

Given that, f(x) = x3/3 – 2âˆšx + 5/x2

Now, differentiate w.r.t. x, we get

â‡’ d(x3/3 – 2âˆšx + 5/x2) / dx

â‡’ 1.d(x3)/3dx – 2d(âˆšx)/dx + 5d(x-2)/dx

â‡’ 1/3.3x2 – 2.1/2.1/âˆšx + 5(-2) x-3

â‡’ x2 – x-1/2 – 10x-3

â‡’ x2 – 1/âˆšx – 10/x3

### Question 4. Differentiate f(x) = exloga + ealogx + ealoga with respect to x.

Solution:

Given that, f(x) = exloga + ealogx + ealoga

Now, differentiate w.r.t. x, we get

â‡’ d(exloga + ealogx + ealoga)

â‡’ d(exloga)/dx + d(ealogx)/dx + d(ealoga)/dx

â‡’ exloga.loga + ealogx.a/x + 0       [As we know, ealoga is constant]

â‡’ loga.exloga + a/x.ealogx

â‡’ loga.ax + a/x xa                [Here, ax can be written as a exloga]

â‡’ ax loga + axa-1

### Question 5. Differentiate f(x) = (2x2 + 1)(3x + 2) with respect to x.

Solution:

Given that, f(x) = (2x2 + 1)(3x + 2)

Now, differentiate w.r.t. x, we get

â‡’ d(2x2 + 1)(3x + 2)/dx

â‡’ (3x + 2)d(2x2 + 1)/dx + (2x2 + 1)d(3x + 2)/dx

â‡’ (3x + 2)(4x+0) + (2x2 + 1)(3+ 0)

â‡’ (12x2 + 8x + 6x2 + 3)

â‡’ 18x2 + 8x + 3.

### Question 6. Differentiate f(x) = log3x + 3logex + 2tanx with respect x.

Solution:

Given that, f(x) = log3x + 3logex + 2tanx

Now, differentiate w.r.t. x, we get

â‡’ d( log3x + 3logex + 2tanx)/dx

â‡’ 1/log3 d(logx)/dx + 3.d(logex)/dx + 2.d(tanx)/dx

â‡’ 1/log3 Ã— 1/x + 3/x + 2sec2x

â‡’ 1/xlog3 + 3/x + 2sec2x

### Question 7. Differentiate f(x) = (x + 1/x) (âˆšx + 1/âˆšx) with respect to x.

Solution:

Given that,  f(x) = (x + 1/x) (âˆšx + 1/âˆšx)

Now, differentiate w.r.t. x, we get

â‡’ d((x + 1/x) (âˆšx + 1/âˆšx))/dx

â‡’ (x + 1/x) d(âˆšx + 1/âˆšx)/dx  + (âˆšx + 1/âˆšx) d(x + 1/x)/dx

â‡’ (x + 1/x) (1/2âˆšx – 1/2x3/2) +  (âˆšx + 1/âˆšx) (1 – 1/x2)

â‡’ {x/(2âˆšx) – x/(2x3/2) + 1/2(x3/2)- 1/(2x5/2)} + {âˆšx – âˆšx/x2 + 1/âˆšx – 1/x5/2}

â‡’ (1.âˆšx/2 – 1/2âˆšx + 1/2x3/2 – 1/2x5/2 + âˆšx – 1/x3/2 + 1/âˆšx – 1/x5/2)

â‡’ (3âˆšx/2 + âˆšx/2 – 1/2x3/2 – 3/2x5/2)

â‡’ 3x1/2/2 + x-1/2/2 – x-3/2/2 – 3x-5/2/2

### Question 8. Differentiate f(x) = (âˆšx + 1/âˆšx)3 with respect to x.

Solution:

Given that, f(x) = (âˆšx + 1/âˆšx)3

Now, differentiate w.r.t. x, we get

â‡’ d(âˆšx + 1/âˆšx)3 /dx

â‡’ d(x3/2 + 3x.1/x + 3âˆšx.1/x + 1/x3/2)/dx      [As we know that, (a + b)3 = a2+ 3a2b + 3ab2 + b3]

â‡’ d(x3/2 + 3x1/2 + 3x-1/2 + x-3/2)/dx

â‡’ 3x1/2/2 + 3x-1/2/2 + 3.(-1/2).x-3/2 – 3x-5/2/2

â‡’ 3x1/2/2 – 3x-5/2/2 + 3x-1/2/2 – 3x-3/2/2.

### Question 9. Differentiate f(x) = 2x2 + 3x + 4 /x with respect to x.

Solution:

Given that, f(x) = 2x2 + 3x + 4 /x

Now, differentiate w.r.t. x, we get

â‡’ d(2x2 + 3x + 4 /x) / dx

â‡’d(2x2/x + 3x/x + 4/x) /dx

â‡’ d(2x + 3 + 4x-1) / dx

â‡’ 2- 4/x2

### Question 10. Differentiate f(x) = (x3+ 1) (x – 2) / x2 with respect x.

Solution:

Given that, f(x) = (x3+ 1) (x – 2) / x2

Now, differentiate w.r.t. x, we get

â‡’ d{(x3 + 1) (x – 2) / x2} / dx

â‡’ d{(x4 – 2x3 +x – 2)/ x2} / dx

â‡’ d(x2 – 2x + x-1 – 2x-2) / dx

â‡’ d(x2)/dx – 2d(x)/dx + d(x-1)/dx – 2d(x-2)/dx

â‡’ 2x – 2 – 1/x2 + 4/x3

â‡’ 2x – 2 – 1/x2 + 4/x3

### Question 11. Differentiate f(x) = acosx + bsinx + c / sinx with respect to x.

Solution:

Given that, f(x) = acosx + bsinx + c / sinx

Now, differentiate w.r.t. x, we get

â‡’ d(acosx + bsinx + c / sinx) /dx

â‡’ a.d(cosx)/dx(sinx) + b.d(1)/dx + c.d/dx(sinx)

â‡’ a(-cosec2x) + 0 + c(-cosecx.cotx)

â‡’ -acosec2x – c.cosecx.cotx

### Question 12. Differentiate f(x) = (2secx + 3cotx – 4tanx) with respect to x.

Solution:

Given that, f(x) = (2secx + 3cotx – 4tanx)

Now, differentiate w.r.t. x, we get

â‡’ d(2secx + 3cotx – 4tanx) / dx

â‡’ 2.d(2secx)/dx + 3.d(cotx)/dx – 4.d(tanx)/dx

â‡’ 2secxtanx – 3cosec2x – 4sec2x

### Question 13. Differentiate f(x) = (a0xn + a1xn-1 + a2xn-2  + ……… + an-1x + an) with respect to x.

Solution:

Given that, f(x) = (a0xn + a1xn-1 + a2xn-2  + ……… + an-1x + an)

Now, differentiate w.r.t. x, we get

â‡’ d(a0xn + a1xn-1 + a2xn-2 + ……… + an-1x + an) / dx

â‡’ a0d(x)n/dx + a1d(x)n-1/dx + a2d(x)n-2/dx + ………. + an-1d(x)/dx + and(1)/dx

â‡’ na0xn-1 + (n-1)a1xn-2 + ………. + an-1 + 0

â‡’ na0xn-1+ (n-1)a1xn-2 + ……….. + an-1

### Question 14. Differentiate f(x) = 1/sinx + 2x+3 + 4/logx3 with respect to x.

Solution:

Given that, f(x) = 1/sinx + 2x+3 + 4/logx

Now, differentiate w.r.t. x, we get

â‡’ d/dx (1/sinx + 2x+3 + 4/logx3)

â‡’ d(cosecx)/dx + 23d(2x)/dx + 4/log3 Ã— d(logx)/dx        [As we know that, logba = loga/logb]

â‡’ -cosecx.cotx + 8 Ã— 2log2 + 4/log3 Ã— 1/x           [Since, d(ax)/dx = axloga]

â‡’ -cosecx.cotx + 2x+3log2 + 4/xlog3

### Question 15. Differentiate f(x) = (x + 5)(2x – 1) / x with respect to x.

Solution:

Given that, f(x) = (x + 5)(2x – 1) / x

Now, differentiate w.r.t. x, we get

â‡’ d/dx {(x + 5)(2x2 – 1)/x}

â‡’ d/dx (2x3 + 10x2 – x – 5 / x)

â‡’ d(2x2 + 10x – 1 – 5x-1)/dx

â‡’ 2d(x2)/dx + 10d(x)/dx – d(1)/dx – 5d(x-1)/dx

â‡’ 2.2x + 10 – 0 + 5/x2

â‡’ 4x + 10 + 5/x

### Question 16. Differentiate f(x) = log(1/âˆšx) + 5xa – 3ax + 3âˆšx2 + 6(4âˆšx-3) with respect to x.

Solution:

Given that, f(x) = log(1/âˆšx) + 5xa – 3ax + 3âˆšx2 + 6(4âˆšx-3

Now, differentiate w.r.t. x, we get

â‡’ d/dx {log(1/âˆšx) + 5xa – 3ax + 3âˆšx2 + 6(4âˆšx-3)}

â‡’ d(log(1/âˆšx)/dx + 5d(xa)/dx – 3(ax) + d(3âˆšx2)/dx + 6d(4âˆšx-3)/dx

â‡’ -1/2.1/x + 5axa-1 – 3axloga + 2x-1/3/3 + 6x-7/4(-3/4)

â‡’ -1/2x + 5axa-1 – 3axloga + 2x-1/3/3 – 9x-7/4/2

### Question 17. Differentiate f(x) = cos(x + a) with respect to x.

Solution:

Given that, f(x) = cos(x + a)

Now, differentiate w.r.t. x, we get

â‡’ d{cos(x + a)}/dx

â‡’ d(cosx.cosa – sinx.sina)/dx

â‡’ cosa.d(cosx)/dx – sina.d(sinx)/dx

â‡’ cosa(-sinx) – sina(cosx)

â‡’ cosx.sina + sinx.cosa

â‡’ -(sinx.cosa + cosx.sina)

â‡’ -sin(x + a)

### Question 18. Differentiate f(x) = cos(x – 2)/sinx with respect to x.

Solution:

Given that, f(x) = cos(x – 2)/sinx

Now, differentiate w.r.t. x, we get

â‡’ d{cos(x – 2)/sinx)/dx

â‡’ d{(cosx.cos2 + sinx.sin2)/sinx} / dx

â‡’ cos2.d(cotx)/dx + sin2.d(1)/dx

â‡’ -cos2.cosec2x + 0

â‡’ -cosec2x.cos2

### Question 19. If y = {sin(x/2) + cos(x/2)}, find dy/dx at x = Ï€/6.

Solution:

Given that, y = {sin(x/2) + cos(x/2)}  …..(1)

Find that dy/dx at x = Ï€/6

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d{sin(x/2) + cos(x/2)}/dx

â‡’ d{sin2(x/2) + cos2(x/2) + 2sin(x/2).cos(x/2)}/dx

â‡’ d(1 + sinx)/dx       [As we know that sin2x + cos2x = 1]

â‡’ 0 + cosx             [As we know that sin2x= 2sinx.cosx]

â‡’ cosx

Now put x = Ï€/6

â‡’ cos(Ï€/6)

â‡’ âˆš3/2

### Question 20. If y = (2 – 3cosx / sinx), find dy/dx at x = Ï€/4.

Solution:

Given that, y = (2 – 3cosx / sinx)  ….(1)

Find that dy/dx at x = Ï€/4

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d(2 – 3cosx / sinx) / dx

â‡’ d(2cosecx – 3cotx) / dx

â‡’ 2d(cosecx)/dx – 3d(cotx)/dx

â‡’ -2cosecx.cotx + 3cosec2x

Now put x = Ï€/4

â‡’ -2cosec(Ï€/4).cot(Ï€/4) + 3cosec2(Ï€/4)

â‡’ -2âˆš2 – 1 + 3.2

â‡’ -2âˆš2 + 6

â‡’ 6 – 2âˆš2

### Question 21. Find the slope of the tangent to the curve f(x) = 2x6 + x4 – 1 at x = 1.

Solution:

Given that f(x) = 2x6 + x4 – 1 at x = 1.

Find the slope of the tangent at a point x = 1

Now, differentiate w.r.t. x, we get

â‡’ d(2x6 + x4 -1)/dx

â‡’ 2dx6/dx + dx4/dx – d.1/dx

â‡’ 12x5+ 4x3 – 0

â‡’ 12x5 + 4x3

Now put x = 1

â‡’ 12(1)5 + 4(1)3

â‡’ 12 + 4

â‡’ 16

Hence, the slope of the tangent to the curve f(x) at x = 1 is 16.

### Question 22. If y = âˆšx/a + âˆša/x, prove that 2xy.dy/dx = (x/a – a/x)

Solution:

Given that, y = âˆšx/a + âˆša/x

Prove that 2xy.dy/dx = (x/a – a/x)

Proof:

dy/dx = d(âˆšx/a + âˆša/x)/dx

â‡’ 1/âˆša.d(âˆšx)/dx + âˆša.d(1/âˆšx)/dx

â‡’ 1/âˆša.1/2âˆšx + âˆša(-1/2).1/xâˆšx

â‡’ 1/2x{âˆšx/a + (-âˆša/x)}

â‡’ 2x.dy/dx = âˆšx/a – âˆša/x

Multiplying both side by y = âˆšx/a + âˆša/x, we get

â‡’ 2xy.dy/dx = (âˆšx/a – âˆša/x)(âˆšx/a + âˆša/x)

â‡’ (x/a – a/x)

Hence proved.

### Question 23. Find the rate at which function f(x) = x4 – 2x3 + 3x2 + x + 5 changes with respect to x.

Solution:

Given that, f(x) = x4 – 2x3 + 3x2 + x + 5

Now, differentiate w.r.t. x, we get

df(x)/dx = d(x4 – 2x3 + 3x2 + x + 5) / dx

â‡’ 4x3 – 6x2 + 6x + 1.

### Question 24. If y = 2x9/3 – 5x7/7 + 6x3 – x, find dy/dx at x = 1.

Solution:

Given that, y = 2x9/3 – 5x7/7 + 6x3 – x   …..(1)

Find that dy/dx at x = 1

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d(2x9/3 – 5x7/7 + 6x3 – x) / dx

â‡’ 2/3dx9/dx – 5/7dx7/dx + 6dx3/dx – dx/dx

â‡’ 2/3.9x8 – 5/7.7x6 + 18x2 – 1

â‡’ 6x8 – 5x6 + 18x2 – 1.

Put x = 1

â‡’ 6(1)8 – 5(1)6 + 18(1)2 – 1

â‡’ 6 – 5 + 18 – 1

â‡’ 18

### Question 25. If f(x) = Î»x2 + Î¼x + 12, f'(4) = 15 and f'(2) = 11, then find Î» and Î¼.

Solution:

Given that, f(x) = Î»x2 + Î¼x + 12  …..(1)

f'(4) = 15 and f'(2) = 11

Find: the value of Î» and Î¼.

Now, differentiate eq(1) w.r.t. x, we get

f(x) = Î»x2 +Î¼x + 12

f'(x) = 2Î»x + Î¼

Now put f'(4) = 15, we get

â‡’ 2Î»(4) + Î¼ = 15

â‡’ 8Î» + Î¼ = 15   ……………(1)

Now put, f'(2) = 11

â‡’ 2Î»(2) + Î¼ = 11

4Î» + Î¼ = 11     …………… (2)

From equation (1) and (2), we get

â‡’ 4Î» = 4

â‡’ Î» = 1

Now put value of Î» in equation (1), we get

â‡’ 8(1) + Î¼ = 15

â‡’ Î¼ = 7

Hence, the value of Î» = 1 and Î¼ = 7

â€‹

### Prove that f'(1) = 100f'(0).

Solution:

Given that, f(x) = x100/100 + x99/99 + ………….. + x2/2 + x + 1

Now, differentiate w.r.t. x, we get

â‡’ f'(x) = x99 + x98 + ………… + x + 1 + 0  ……………..(1)

From equation (1),

â‡’ f'(1) = 1 + 1 + …………….(100 times)

â‡’ 100

Again,

â‡’ f'(0) = 0 + 0 + ………….. + 1

â‡’ 1

Now,

â‡’ f'(1) = 100

â‡’ 100 Ã— 1 = 100 Ã— f'(0)

â‡’ f'(1) = 100f'(0)

Hence Proved

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