Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type to help strengthen their preparations for exams.
TS Inter 1st Year Maths 1B Differentiation Important Questions Long Answer Type
Question 1.
If y = \(\tan ^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]\), then find \(\frac{\mathbf{d y}}{\mathbf{d x}}\). [Mar. ’18 (TS); Mar. ’16 (AP), ’12, ’10, ’09, ’04; May ’15 (AP & TS), ’12, ’97]
Solution:
Question 2.
If y = xtan x + (sin x)cos x, then find \(\frac{\mathbf{d y}}{\mathbf{d x}}\). [Mar. ’14, ’13 (Old), ’11, ’08, ’07; May ’13, ’06; Mar. ’18 (AP)]
Solution:
Given y = xtan x + (sin x)cos x
Differentiating on both sides with respect to ‘x’.
Let u = xtan x
v = sin xcos x
y = u + v
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{u}+\mathrm{v})\)
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) ……(1)
Now, u = xtan x
Taking logarithms on both sides,
log u = log xtan x
log u = tan x log x
Differentiating on both sides with respect to ‘x’.
Now, v = (sin x)cos x
Taking logarithms on both sides,
log v = log (sin x)cos x
log v = cos x log(sin x)
Differentiating on both sides with respect to ‘x’.
Question 3.
If x = \(\frac{3 a t}{1+t^3}\), y = \(\frac{3 \mathrm{at}^2}{1+\mathrm{t}^3}\), then find \(\frac{\mathbf{d y}}{\mathbf{d x}}\). [B.P.]
Solution:
Given that x = \(\frac{3 a t}{1+t^3}\)
Differentiating on both sides with respect to ‘t’.
Question 4.
If \(\sqrt{1-x^2}+\sqrt{1-y^2}\) = a(x – y) then show that \(\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}\). [Mar. ’17 (TS), ’08, ’05; May ’14, ’13 (Old), ’11, ’97]
Solution:
Given that \(\sqrt{1-x^2}+\sqrt{1-y^2}\) = a(x – y)
Put x = sin α ⇒ α = sin-1x
y = sin β ⇒ β = sin-1y
Now, \(\sqrt{1-\sin ^2 \alpha}+\sqrt{1-\sin ^2 \beta}\) = a(sin α – sin β)
\(\sqrt{\cos ^2 \alpha}+\sqrt{\cos ^2 \beta}\) = a(sin α – sin β)
cos α + cos β = a(sin α – sin β)
Question 5.
If y = \(\mathbf{x} \sqrt{\mathbf{a}^2+x^2}+a^2 \log \left(x+\sqrt{a^2+x^2}\right)\) then show that \(\frac{d y}{d x}=2 \sqrt{a^2+x^2}\). [Mar. ’19, ’15 (AP). ’09, ’02; May ’08]
Solution:
Given, that y = \(\mathbf{x} \sqrt{\mathbf{a}^2+x^2}+a^2 \log \left(x+\sqrt{a^2+x^2}\right)\)
Differentiating on both sides with respect to ‘x’.
Question 6.
If y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)\) – \(\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)\) then show that \(\frac{\mathbf{d y}}{\mathbf{d x}}=\frac{1}{1+x^2}\). [May ’07]
Solution:
Given that
y = 2θ + 3θ – 4θ = θ
y = tan-1x
Differentiating on both sides with respect to x.
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \tan ^{-1} \mathrm{x}\)
∴ \(\frac{d y}{d x}=\frac{1}{1+x^2}\)
Question 7.
If y = \(\frac{(1-2 x)^{2 / 3}(1+3 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}}\), then find \(\frac{\mathbf{d y}}{\mathbf{d x}}\). [May ’10]
Solution:
Question 8.
If y = (sin x)log x + xsin x then find \(\frac{\mathbf{d y}}{\mathbf{d x}}\). [Mar. ’17 (AP), ’15 (TS), ’13]
Solution:
Given that, let y = (sin x)log x + xsin x
Let, u = (sin x)log x , v = xsin x then y = u + v
Differentiating on both sides with respect to x.
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) …….(1)
Now, u = (sin x)log x
Taking logarithms on both sides
log u = log (sin x)log x
log u = log x . log (sin x)
Differentiating on both sides with respect to ‘x’.
Now, v = xsin x
Taking logarithms on both sides
log v = log xsin x
log v = sin x log x
Differentiating on both sides with respect to ‘x’.
Question 9.
If xy + yx = ab then show that \(\frac{d y}{d x}=-\left[\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}\right]\). [Mar. ’03; Mar. ’16 (TS)]
Solution:
Given that, xy + yx = ab
Let, xy = u, yx = v then, u + v = ab
Differentiating on both sides with respect to x.
\(\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{dv}}{\mathrm{dx}}=0\) ……(1)
Now, u = xy
Taking logarithms on both sides
log u = log xy
log u = y log x
Differentiating on both sides with respect to ‘x’.
Now, v = yx
Taking logarithms on both sides
log v = log yx
log v = x log y
Differentiating on both sides with respect to ‘x’.
Question 10.
If f(x) = \(\sin ^{-1} \sqrt{\frac{x-\beta}{\alpha-\beta}}\) and g(x) = \(\tan ^{-1} \sqrt{\frac{x-\beta}{\alpha-x}}\) then show that f'(x) = g'(x), (β < x < α).
Solution:
Some More Maths 1B Differentiation Important Questions Long Answer Type
Question 11.
Find the derivative of 20log(tan x).
Solution:
Let y = 20log(tan x)
Differentiating on both sides with respect to ‘x’.
Question 12.
If f(x) = e2x log x (x > 0), then find f'(x).
Solution:
f(x) = e2x (log x)
Let y = e2x (log x)
Differentiating with respect to ‘x’ on both sides,
Question 13.
If y = \(\frac{\mathbf{a}-\mathbf{x}}{\mathbf{a}+\mathbf{x}}\), find \(\frac{\mathbf{d y}}{\mathbf{d x}}\).
Solution:
Given y = \(\frac{\mathbf{a}-\mathbf{x}}{\mathbf{a}+\mathbf{x}}\)
Differentiating with respect to x on both sides.
Question 14.
If y = sin-1(cos x) then find \(\frac{d y}{d x}\).
Solution:
Let y = sin-1(cos x)
Differentiating on both sides with respect to ‘x’.
Question 15.
If x4 + y4 – a2xy = 0, find \(\frac{\mathbf{d y}}{\mathbf{d x}}\).
Solution:
Given that x4 + y4 – a2xy = 0
Differentiating on both sides with respect to ‘x’.
Question 16.
Find the derivative of cos-1(4x3 – 3x) with respect to ‘x’.
Solution:
Let y = cos-1(4x3 – 3x)
Put x = cos θ
⇒ θ = cos-1x
Now, y = cos-1(4 cos3θ – 3 cos θ)
= cos-1(cos 3θ)
= 3θ
y = 3 cos-1x
Differentiating on both sides with respect to ‘x’.
Question 17.
Find the derivative of sec x from the first principle.
Solution:
Given, f(x) = sec x
Now, f(x + h) = sec (x + h)
Question 18.
Find the derivative of \(\sin ^{-1}\left(\frac{b+a \sin x}{a+b \sin x}\right)\).
Solution:
Let y = \(\sin ^{-1}\left(\frac{b+a \sin x}{a+b \sin x}\right)\)
Differentiating on both sides with respect to ‘x’.
Question 19.
Find the derivative of \(\tan ^{-1}\left(\frac{\cos x}{1+\cos x}\right)\)
Solution:
Let y = \(\tan ^{-1}\left(\frac{\cos x}{1+\cos x}\right)\)
Differentiating on both sides with respect to ‘x’.
Question 20.
Find the derivative of \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to tan-1x. [May ’09]
Solution:
Question 21.
If f(x) = x ex sin x, then find f'(x).
Solution:
f(x) = x . ex . sin x
Let y = x . ex . sin x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d}{d x}\) (x . ex . sin x)
= x . ex . \(\frac{d}{d x}\) (sin x) + x . sin x . \(\frac{d}{d x}\) (ex) + sin x . ex . \(\frac{d}{d x}\) (x)
= x . ex . cos x + x . sin x . ex + sin x . ex (1)
= ex (x cos x + x sin x + sin x)
∴ f'(x) = ex (x cos x + x sin x + sin x)
Question 22.
If f(x) = sin(log x), (x > 0), then find f'(x). [Mar. ’18 (AP)]
Solution:
f(x) = sin(log x)
Let y = sin(log x)
Question 23.
If f(x) = (x3 + 6x2 + 12x – 13)100, then find f'(x).
Solution:
Given that, f(x) = (x3 + 6x2 + 12x – 13)100
Let y = (x3 + 6x2 + 12x – 13)100
Differentiating with respect to x on both sides.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d}{d x}\) (x3 + 6x2 + 12x – 13)100
= 100(x3 + 6x2 + 12x – 13)100-1 \(\frac{d}{d x}\)(x3 + 6x2 + 12x – 13)
= 100(x3 + 6x2 + 12x – 13)99 (3x2 + 6(2x) + 12(1) – 0)
= 100(x3 + 6x2 + 12x – 13)99 (3x2 + 12x + 12)
= 100(x3 + 6x2 + 12x – 13)99 3(x2 + 4x + 4)
= 300(x + 2)2 (x3 + 6x2 + 12x – 13)99
∴ f'(x) = 300(x + 2)2 . (x3 + 6x2 + 12x – 13)99
Question 24.
Find the derivative of (ax + b)n (cx + d)m.
Solution:
Given, f(x) = (ax + b)n (cx + d)m
Let y = (ax + b)n (cx + d)m
Differentiating with respect to ‘x’ on both sides.
Question 25.
Find the derivative of \(\frac{\mathbf{p} \mathbf{x}^2+\mathbf{q x}+\mathbf{r}}{\mathbf{a x}+\mathbf{b}}\).
Solution:
Given, f(x) = \(\frac{\mathbf{p} \mathbf{x}^2+\mathbf{q x}+\mathbf{r}}{\mathbf{a x}+\mathbf{b}}\)
Let y = \(\frac{\mathbf{p} \mathbf{x}^2+\mathbf{q x}+\mathbf{r}}{\mathbf{a x}+\mathbf{b}}\)
Differentiating with respect to ‘x’ on both sides.
Question 26.
Find the derivative of \(\log _7(\log x)\).
Solution:
Given that, f(x) = \(\log _7(\log x)\)
Let y = \(\log _7(\log x)\)
Differentiating with respect to ‘x’ on both sides.
Question 27.
Find the derivative of the function f(x) = (x2 – 3)(4x3 + 1). [May ’15 (AP)]
Solution:
Question 28.
Find the derivative of tan-1(log x). [Mar. ’19 (AP); May ’15 (TS)]
Solution:
Question 29.
If f(x) = 2x3 + 3x – 5, then prove that f'(0) + 3 . f'(-1) = 0. [Mar. ’16 (AP)]
Solution:
Given f(x) = 2x2 + 3x – 5
Now f'(x) = 2(2x) + 3(1) – 0 = 4x + 3
f'(0) = 4(0) + 3 = 3
f(-1) = 4(-1) + 3 = -4 + 3 = -1
LHS = f'(0) + 3. f'(-1)
= 3 + 3(-1)
= 3 – 3
= 0
∴ f'(0) + 3 . f'(-1) = 0
Question 30.
If 2x2 – 3xy + y2 + x + 2y – 8 = 0, then find \(\frac{d \mathbf{y}}{\mathbf{d x}}\). [Mar. ’16 (TS)]
Solution:
Given 2x2 – 3xy + y2 + 2y – 8 = 0
differentiating on both sides with respect to ‘x’.
Question 31.
If ay4 = (x + b)5 then 5yy11 = (y1)2. [Mar. ’17 (TS)]
Solution:
Question 32.
If y = x4 + tan x then find y11. [Mar. ’18 (AP)]
Solution:
Given that y = x4 + tan x
y1 = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x4 + tan x) = 4x3 + sec2x
y11 = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(4x3 + sec2x)
= 4(3x2) + 2 sec x \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sec x)
= 12x2 + 2 sec x (sec x tan x)
∴ y11 = 12x2 + 2 sec2x tan x
Question 33.
If y = \(\frac{2 x+3}{4 x+5}\), then find y”.
Solution:
y = \(\frac{2 x+3}{4 x+5}\)
Question 34.
If f(x) = log(tan ex), then find f'(x). [Mar. ’19 (TS)]
Solution:
Question 35.
Evaluate \({Lim}_{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}\). [Mar. ’19 (TS)]
Solution:
Question 36.
If xlog y = log x, then show that \(\frac{d y}{d x}=\frac{y}{x}\left(\frac{1-\log x \log y}{(\log x)^2}\right)\). [Mar. ’19 (TS)]
Solution:
