OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Ex 8(l)

Well-structured S Chand ISC Maths Class 12 Solutions Chapter 8 Differentiation Ex 8(l) facilitate a deeper understanding of mathematical principles.

S Chand Class 12 ICSE Maths Solutions Chapter 8 Differentiation Ex 8(l)

Find the second derivative of the following functions :

Question 1.
(i) x²
(ii) a^x
(iii) ax³ + bx² + cx + d
(iv) log x
(v) 1/\(\sqrt{x}\)
(vi) x/\(\sqrt{x-1}\)
(vii) sin^-1 x
Solution:
(i) Let y = x² ; Diff. both sides w.r.t. x,
\(\frac { dy }{ dx }\) = 2x ; Again diff. both sides w.r.t. x
∴ \(\frac{d^2 y}{d x^2}\) = 2

(ii) Let y = a^x ; Diff. both sides w.r.t. x, we have
\(\frac { dy }{ dx }\) = a^x log a; Again diff. both sides w.r.t. x ; we have
∴ \(\frac{d^2 y}{d x^2}\) = a²(log a)²

(iii) Let y = ox³ + bx² + cx + d; Diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = 3ax² + 2bx + c; Diff. again w.r.t. x
\(\frac { d²y }{ dx² }\) = 6ax + 2b

(iv) Let y = log x ; Diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\) ; Diff. again w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=-\frac{1}{x^2}\)

Question 2.
(i) e^x + sin x
(ii) e^-x sin x
Solution:
(i) Let y = e^x + sin x ;
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = e^x + cos x ;
Diff. again both sides w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=e^x-\sin x\)

(ii) Let y = e^-x sin x ;
Diff. both sides w.r.t. x,

Question 3.
(i) If y = 2 sin x + 3 cos x, prove that y + \(\frac{d^2 y}{d x^2}\) = 0.
(ii) If y = a + bx², prove that x.\(\frac{d^2 y}{d x^2}=\frac{d y}{d x}\)
(iii) If y = tan x + sec x, prove that \(\frac{d^2 y}{d x^2}=\frac{\cos x}{(1-\sin x)^2}\).
(iv) If y = 500, e^7x + 600 e^-7x, show that \(\frac{d^2 y}{d x^2}\) = 49 y.
(iv) If e^y (1 + x) = 1, show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\).
Solution:
(i) Given y = 2 sin x + 3 cos x …(1)
Diff. both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 2 cos x – 3 sin x ;
Again diff. both sides w.r.t. x

(iii) Given y = tan x + sec x … (1)
Diff. eqn. (1) both sides w.r.t. x; we have

(iv) Given y = 500 e^7x + 600 e^-7x …(1)
Diff. eqn. (1) both sides w.r.t. x; we have
\(\frac { dy }{ dx }\) = 3500 e^7x – 4200 e^-7x
Again diff. both sides w.r.t. x
\(\frac { dy }{ dx }\) = 7 x 3500 e^7x + 4200 x 7 e^-7x
= 49[500 e^7x + 600 e^-7x] = 49 y [using eqn. (1)]

(v) Given e^y (1 + x) = 1 ⇒ e^y = \(\frac { 1 }{ 1 + x }\)
Taking logorithm both sides w.r.t. x, we have
y = log\(\left(\frac{1}{1+x}\right)\) = – log(1 + x)
Diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = – \(\left(\frac{1}{1+x}\right)\) … (1)
Again diff. both sides w.r.t. x
\(\frac{d^2 y}{d x^2}=\frac{1}{(1+x)^2}=\left(\frac{d y}{d x}\right)^2\) [using eqn. (1)]

Question 4.
If y = tan x, prove that \(\frac{d^2 y}{d x^2}=2 y \frac{d y}{d x}\).
Solution:

Question 5.
If y = \(\frac{\log x}{x}\), prove that \(\frac{d^2 y}{d x^2}=\frac{2 \log x-3}{x^3}\).
Solution:
Given y = \(\frac{\log x}{x}\)
Diff. both sides w.r.t. x; we have

Question 6.
(i) If y = tan^-1 x, prove that
(1 + x²) \(\frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}\) = 0.
(ii) If y = sin^-1x, then show that
(1 + x²) \(\frac{d^2 y}{d x^2}-x \frac{d y}{d x}\) = 0.
Solution:
(i) Given y = tan^-1 x;
Diff. both sides w.r.t. x, we have

(ii) Given y = sin^-1 x;
Diff. both sides w.r.t. x, we have

Question 7.
If y = \(e^{\tan ^{-1} x}\), prove that
\(\left(1+x^2\right) \frac{d^2 y}{d x^2}+(2 x-1) \frac{d y}{d x}\) = 0.
Solution:

Question 8.
If y = x^x, prove that
\(\frac{d^2 y}{d x^2}-\frac{1}{y}\left(\frac{d y}{d x}\right)^2-\frac{y}{x}\) = 0
Solution:
Given y = x^x, … (1)
Taking logarithm on eqn. (1); we have
log y = x log x …(2)
Diff. eqn. (2) w.r.t. x, we have

Question 9.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\), prove that
\(\left(1-x^2\right) \frac{d^2 y}{d x^2}-3 x \frac{d y}{d x}-y\) = 0.
Solution:

Question 10.
If y = (tan^-1 x)², prove that
\(\left(x^2+1\right)^2 \frac{d^2 y}{d x^2}+2 x\left(x^2+1\right) \frac{d y}{d x}\) = 2.
Solution:
Given y = (tan^-1x)²,
Diff. both sides w.r.t. x, we have

Question 11.
If y = sin (m sin^-1 x) show that (1 – x²)\(\frac{d^2 y}{d x^2}-x \frac{d y}{d x}+m^2 y\)y = 0
Solution:
Given y = sin (m sin^-1x) … (1)
Diff. eqn (1) w.r.t. x, we have

Question 12.
If y = (A + Bx)e^3x, prove that y” + 6y’ + 9y + 2 = 2.
Solution:
Given y = (A + Bx)e^-3x …(1)
Diff. eqn. (1) w.r.t. x, we have

Question 13.
If x^myⁿ = (x + y)^m+n, prove that \(\frac{d^2 y}{d x^2}\) = 0.
Solution:
Given x^myⁿ = (x + y)^m+n
Taking logaritum on both sides, we have

Question 14.
If y = ae^mx + be^-mx, prove that \(\frac{d^2 y}{d x^2}-m^2 y\) = 0.
Solution:

Question 15.
If y = a cos (log x) + b sin (log x), prove that \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y\) = 0.
Solution:

Question 16.
Find \(\frac{d^2 y}{d x^2}\) when
(i) x = t², y = t³.
(ii) x = at², y = 2at.
(iii) x = a cos θ, y = b sin θ
(iv) x = cos t, y = sin t
Solution:
(i) Let x = t² … (1)
& y = t³ … (2)
Diff. eqn. (1) & (2) w.r.t. t; we have

(ii) x = at² … (1)
& y = 2at … (2)

(iii) Let x = a cos θ …(1)
& y = b sin θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Question 17.
Find \(\frac{d^2 y}{d x^2}\) when θ = \(\frac { π }{ 2 }\):
(i) x = a(θ + sin θ), y = a(1 – cos θ)
(ii) x = a(1 – cos θ), y = a(θ + sin θ).
Solution:
(i) Let x = a(θ + sin θ) …(1)
& y = a(1 – cos θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

(ii) Given x = a(1 – cos θ) …(1)
& y = a(θ + sin θ) …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Question 18.
If x = a sec³θ, y = a tan³θ, find \(\frac{d^2 y}{d x^2}\) at θ = \(\frac { π }{ 4 }\).
Solution:
Let x = a sec³θ …(1)
& y = a tan³θ …(2)
Diff. eqn. (1) & (2) w.r.t. θ; we have

Question 19.
If x = cos θ + θ sin θ, y = sin θ – θ cos θ, 0 < θ < \(\frac { π }{ 2 }\), prove that \(\frac{d^2 y}{d x^2}=\frac{\sec ^3 \theta}{\theta}\).
Solution:

Question 20.
If x = cos θ, y = sin³θ, show that \(\frac{d^2 y}{d x^2} \cdot\left(\frac{d y}{d x}\right)^2=3 \sin ^2 \theta\left(5 \cos ^2 \theta-1\right)\) .
Solution:

Question 21.
If f(x) = \(\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec \theta & \tan x & x \\
1 & \tan x-\tan \theta & 0
\end{array}\right|\), then f'(θ) is
(a) 0
(b) – 1
(c) independent of θ
(d) None of these.
Solution:

Question 22.
If y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Solution:

Examples

Question 1.
If y = \(\log \sqrt{\frac{1-\cos x}{1+\cos x}}, \text { find } \frac{d y}{d x}\).
Solution:

Question 2.
If y = (cos x)^{cos x}, find \(\frac { dy }{ dx }\).
Solution:
Given y = (cos x)^{cos x} ;
Taking logarithm or both sides, we have
log y = log (cos x)^{cos x}
log y = cos x log cos x
Diff both sides w.r.t x ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = cos x \(\frac{1}{\cos x}\) (- sin x) + log cos x (- sinx)
\(\frac { dy }{ dx }\) = y [- sinx – sinx log cos x]
= (cos x)^{cos x} (- sinx) [1 + log cos x]
= – sinx (cos x)^{cos x} [log e + log cos x]
= – sinx (cos x)^{cos x} log (e cos x)
[∵ log a + log b = log ab]

Question 3.
If y = e^x log (tan 2x), find \(\frac { dy }{ dx }\).
Solution:
Given y = e^x log (tan 2x) ;
Diff. both sides w.r.t x ; we have
\(\frac { dy }{ dx }\) = e^x \(\frac{1}{\tan 2 x}\) sec²2x.2 + log (tan 2x).e^x
= e^x [2cot 2x sec²2x + log (tan 2x)]

Question 4.
If y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\), prove that \(\frac{d y}{d x}=\frac{2}{1+x^2}\).
Solution:
Given y = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) ;
put x = tan θ ⇒ θ = tan^-1x
∴ y = tan^-1\(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\) = tan^-1(tan 2θ)
⇒ y = 2θ = 2 tan^-1 x
DifF. both sides w.r.t x ; we have
\(\frac{d y}{d x}=\frac{2}{1+x^2}\)

Question 5.
If y = \(e^{m \cos ^{-1} x}\), prove that
\(\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y\).
Solution:
Given y = \(e^{m \cos ^{-1} x}\)
Diff both sides w.r.t x ; we have

Question 6.
If x^yy^x = 5, show that \(\frac{d y}{d x}=-\left(\frac{\log y+\frac{y}{x}}{\log x+\frac{x}{y}}\right)\).
Solution:
Given x^yy^x = 5;
Taking logarithm on both sides; we get
log x^y + log y^x = log 5
⇒ y log x + x log y = log 5 :
Diff both sides w.r.t. x ; we get

Question 7.
If x = a sin³t and y = a cos³t, find \(\frac { dy }{ dx }\).
Solution:
Given x = a sin³t; …(1)
y = a cos³t …(2)
Diff eqn (1) & eqn (2) w.r.t. t, we have

Question 8.
If sin (xy) + cos (xy) = 1 and tan (xy) ≠ 1, then show that \(\frac{d y}{d x}=-\frac{y}{x}\).
Solution:
Given sin (xy) + cos (xy) = 1 …(1)
Diff eqn (1) & eqn (2) w.r.t. x, we have

Question 9.
If x^py^q = (x + y)^p+q, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given x^py^q = (x + y)^p+q;
Taking logarithm on both sides; we have
p log x + q log y = (p + q) log(x + y)
Diff both sides w.r.t x; we have

Question 10.
If y = \(e^{\sin \left(x^2\right)}\), find \(\frac { dy }{ dx }\).
Solution:
Given y = \(e^{\sin \left(x^2\right)}\);
Diff both sides w.r.t. x, we have
\(\frac{d y}{d x}=e^{\sin \left(x^2\right)} \cos x^2 \cdot 2 x\)

Question 11.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^2}}\), prove that (1 – x²) \(\frac { dy }{ dx }\) – xy = 1.
Solution:

Question 12.
If e^x+y = xy, show that \(\frac{d y}{d x}=\frac{y(1-x)}{x(y-1)}\).
Solution:
Given e^x+y = xy ;
Taking logarithm on both sides, we have

Question 13.
If sin y = x sin (a + y), show that \(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}\).
Solution:

Question 14.
Find \(\frac { dy }{ dx }\) if y = tan^-1\(\frac{\sqrt{1+x^2}-1}{x}\).
Solution:

Question 15.
If y = \(\sqrt{\frac{1-\cos x}{1+\cos x}}\), find \(\frac { dy }{ dx }\).
Solution:

Question 16.
Using a suitable substitution, find the derivative of tan^-1\(\frac{4 \sqrt{x}}{1-4 x}\) with respect to x
Solution:

Question 17.
Find the derivative of sin x2 with respect to x³.
Solution:
Let y = sin x² & z = x³ …(2)
So we want to diff. y w.r.t. z i.e. to find \(\frac { dy }{ dx }\)
diff. eqn. (1) & eqn. (2) both sales w.r.t. – x ; we have
\(\frac { dy }{ dx }\) = cos x². 2x ; \(\frac { dz }{ dx }\) = 3x²
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{2 x \cos x^2}{3 x^2}=\frac{2 \cos x^2}{3 x}\)

Question 18.
Using a suitable substitution and the derivative of tan^-1\(\sqrt{\frac{a-x}{a+x}}\) with respect to x.
Solution:

Question 19.
If y = x^x, prove that \(\frac{d^2 y}{d x^2}-\frac{1}{y}\left(\frac{d y}{d x}\right)^2-\frac{y}{x}\) = 0.
Solution:
Given y = x^x … (1)
Taking logarithm on eqn. (1); we have log y = x log x …(2)
Diff. eqn. (2) w.r.t. x, we have
\(\frac{1}{y} \frac{d y}{d x}=x \cdot \frac{1}{x}\) + log x = 1 + log x
∴ \(\frac{d y}{d x}=y(1+\log x)\) … (3)
Diff. eqn. (3) both sides w.r.t. x; we have
\(\frac{d^2 y}{d x^2}=\frac{d y}{d x}(1+\log x)+\frac{y}{x}\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{1}{y}\left(\frac{d y}{d x}\right)^2+\frac{y}{x}\)
[using eqn. (3)]

Question 20.
If e^y (x + 1) = 1, then show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\).
Solution:
Given e^y (x + 1) = 1 ⇒ e^y = \(\frac { 1 }{ 1+x }\)
Taking logorithm both sides w.r.t. x, we have

Question 21.
If y = (cot^-1 x)², show that
\(\left(1+x^2\right)^2 \frac{d^2 y}{d x^2}+2 x\left(1+x^2\right) \frac{d y}{d x}\) = 2.
Solution:

Question 22.
If y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}\), prove that (1 – x²) \(\frac{d y}{d x}=x+\frac{y}{x}\).
Solution:

Question 23.
If log y = tan^-1x, prove that
(1 + x²) \(\frac{d^2 y}{d x^2}+(2 x-1) \frac{d y}{d x}\) = 0.
Solution:

Question 24.
If y = cos (sin x), show that \(\frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x\) = 0
Solution:
Given = cos (sin x) …(1)
diff. both sides w.r.t. x ; we have
\(\frac { dy }{ dx }\) = – sin (sin x) cos x …(2)
again diff. both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – [sin (sin x) (- sin x) + cos² x cos (sin x)]
⇒ \(\frac{d^2 y}{d x^2}=-\frac{\sin x}{\cos x} \frac{d y}{d x}-y \cos ^2 x\) [using (1) and (2)]
⇒ \(\frac{d^2 y}{d x^2}+\tan x \frac{d y}{d x}+y \cos ^2 x\) = 0

Question 25.
If y = sec (tan^-1 x), then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{x}{\sqrt{1+x^2}}\)
(b) \(x \sqrt{1+x^2}\)
(c) \(\sqrt{1+x^2}\)
(d) \(\frac{1}{\sqrt{1+x^2}}\)
Solution:
Let y = sec (tan^-1 x)
⇒ y = sec (sec^-1 \(\sqrt{1+x^2}\)) ⇒ y = \(\sqrt{1+x^2}\)
Diff both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{2}\left(1+x^2\right)^{-\frac{1}{2}} \times 2 x=\frac{x}{\sqrt{1+x^2}}\)

Question 26.
Differentiate sin (sin 2x).
(a) 2 cos 2x cos 2x
(b) 2 cos 2x cos (sin 2x)
(c) 2 cos 2x sin 2x
(d) cos 2x cos (sin 2x)
Solution:
Let y = sin (sin 2x)
\(\frac { dy }{ dx }\) = cos (sin 2x) \(\frac { d }{ dx }\) sin 2x = cos (sin 2x) 2 cos 2x

Question 27.
If x = ct and y = \(\frac { c }{ t }\), find \(\frac { dy }{ dx }\), at t = 2.
(a) 4
(b) 0
(c) \(\frac { 1 }{ 4 }\)
(d) – \(\frac { 1 }{ 4 }\)
Solution:
Given x = ct … (1)
and y = \(\frac { c }{ t }\) … (2)
Diff. both eqns. (1) and (2) w.r.t. t, we get

Question 28.
If y = tan^-1\(\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)\), then \(\frac { dy }{ dx }\) is equal to
(a) 0
(b) \(\frac { 1 }{ 2 }\)
(c) \(\frac { π }{ 4 }\)
(d) 1
Solution:

Question 29.
If y = tan^-1 x + cot^-1 x + sec^-1 x + cosec^-1 x, then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{x^2-1}{x^2+1}\)
(b) π
(c) 0
(d) 1
(e) \(\frac{1}{x \sqrt{x^2-1}}\)
Solution:
y = (tan^-1 x + cot^-1 x) + (sec^-1 x + cosec^-1 x)
⇒ y = \(\frac { π }{ 2 }\) + \(\frac { π }{ 2 }\)
∴ \(\frac { dy }{ dx }\) = 0

Question 30.
If y = sin^-1 \(\sqrt{1-x}\), then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac{1}{\sqrt{1-x}}\)
(b) \(\frac{-1}{2 \sqrt{1-x}}\)
(c) \(\frac{1}{\sqrt{x}}\)
(d) \(\frac{-1}{2 \sqrt{x} \sqrt{1-x}}\)
Solution:

Question 31.
If x ≠ 0 and y = log, |2 x|, then \(\frac { dy }{ dx }\) is equal to
(a) \(\frac { 1 }{ x }\)
(b) \(\frac { -1 }{ x }\)
(c) ± \(\frac { 1 }{ 2x }\)
(d) None of these
Solution:
Given y = log | 2x | ; x ≠ 0, when x < 0 then | 2x | = – 2x ∴ y = log (- 2x) ⇒ \(\frac{d y}{d x}=\frac{-1}{2 x}(-2)=\frac{1}{x}\) when x > 0 Then |2x| = 2x
∴ y = log2x
⇒ \(\frac{d y}{d x}=\frac{1}{2 x} \times 2=\frac{1}{x}\)
Thus \(\frac { dy }{ dx }\) = \(\frac { 1 }{ x }\) ; when x ≠ 0

Question 32.
If x² + y² = 4, then y\(\frac { dy }{ dx }\) + x =
(a) 4
(b) 0
(c) 1
(d) – 1
Solution:
Given x² + y² = 4 ; diff. both sides w.r.t. x ;
2x + 2y \(\frac { dy }{ dx }\) = 0 ⇒ x + y\(\frac { dy }{ dx }\) = 0

Question 33.
If y = sin^-1 \(\left(2 x \sqrt{1-x^2}\right), \quad-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}\), then \(\frac { dy }{ dx }\) is equal t0
Solution:

Question 34.
If y = tan^-1 \(\left(\frac{a-x}{1+a x}\right)\), then \(\frac { dy }{ dx }\)
(a) \(\frac{1}{1+x^2}\)
(b) \(\frac{a}{1+a x^2}\)
(c) – \(\frac{1}{1+x^2}\)
(d) \(\frac{x}{1+x^2}\)
Solution:

Question 35.
If y = log\(\left(\frac{x^2}{e^2}\right)\), then \(\frac{d^2 y}{d x^2}\) equals
(a) \(\frac{1}{1+x^2}\)
(b) – \(\frac{a}{1+a x^2}\)
(c) \(\frac{1}{1+x^2}\)
(d) – \(\frac{x}{1+x^2}\)
Solution:

Question 36.
If y = Ae^5x + Be^-5x, then \(\frac{d^2 y}{d x^2}\) is equal to
(a) 25y
(b) 5y
(c) – 25y
(d) 15y
Solution:

Question 37.
If y = | cos x | + | sin x |, then \(\frac { dy }{ dx }\) at x = \(\frac { 2π }{ 3 }\) is
(a) \(\frac{1-\sqrt{3}}{2}\)
(b) 0
(c) \(\frac{1}{2}(\sqrt{3}-1)\)
(d) None of these
Solution:

Find \(\frac { dy }{ dx }\) if y =

Question 38.
cosec x°
Solution:
Given y = cosec x° = cosec \(\frac { πx }{ 180 }\) [π rad = 180°]
∴ \(\frac { dy }{ dx }\) = – cot \(\frac { πx }{ 180 }\) cosec \(\frac { πx }{ 180 }\).\(\frac { π }{ 180 }\) = – \(\frac { π }{ 180 }\) cot x° cosec x°

Question 39.
cos x³
Solution:
Given y = cos x³
∴ \(\frac { dy }{ dx }\) = – sinx³ (3x²)

Question 40.
sin (sin 3x)
Solution:
Given y = sin (sin 3x)
∴ \(\frac { dy }{ dx }\) = cos (sin3x) \(\frac { d }{ dx }\) sin 3x = 3 cos (sin 3x) cos 3x

Question 41.
log (tan x)
Solution:
Given y = log (tan x)
∴ \(\frac { dy }{ dx }\) = ∴ \(\frac{1}{\tan x} \frac{d}{d x} \tan x=\frac{\sec ^2 x}{\tan x}=\frac{1}{\sin x \cos x}\) = sec x. cosec x

Question 42.
xy = c²
Solution:
Given xy = c² ; diff. both sides w.r.t. x
x\(\frac { dy }{ dx }\) + y.1 = 0 ⇒ \(\frac { dy }{ dx }\) = – \(\frac { y }{ x }\)

Question 43.
log (cos e^x)
Solution:
Given y = log (cos e^x)
∴ \(\frac{d y}{d x}=\frac{1}{\cos e^x} \frac{d}{d x} \cos e^x=\frac{1}{\cos e^x}\left\{-\sin e^x\right\} \frac{d}{d x} e^x=-e^x \tan e^x\)

Question 44.
cosec (cot \(\sqrt{x}\))
Solution:

Question 45.
\(\tan ^{-1} \frac{1+\cos x}{\sin x}\)
Solution:

Question 46.
If f(x) = x + 1, then write the value of \(\frac { d }{ dx }\) (fof) (x).
Solution:
Given f(x) = x + 1
∴ (fof) (x) = f(f (x)) = f(x + 1) = x + 1 + 1 = x + 2
∴ \(\frac { d }{ dx }\) (fof) (x) = 1

Question 47.
If f (x) = | cos x |, then f'(\(\frac { π }{ 4 }\)) is equal to …………..
Solution:

Question 48.
If f (x) = x | x |, then f'(x) = ……………
Solution:
Given, f (x) = x | x | ; difF. both sides w.r.t. x ; we have

Question 49.
If f(x) = | cos x – sin x |, then f”(\(\frac { π }{ 3 }\)) = ………….
Solution:

Question 50.
If y = a^x, then find \(\frac{d^2 y}{d x^2}\).
Solution:
Given y = a^x
∴ \(\frac{d y}{d x}=a^x \log a\)
∴ \(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(a^x \log a\right)=\log a \cdot a^x \log a=a^x(\log a)^2\)

Question 51.
For the curve \(\sqrt{x}+\sqrt{y}=1, \frac{d y}{d x} \text { at }\left(\frac{1}{4}, \frac{1}{4}\right)\) is …………
Solution:

Question 52.
Write the derivative of sin x w.r.t. cos x.
Solution:
Let y = sin x
and z = cos x
We want to find \(\frac { dy }{ dz }\)
Diff. both eqns. (1) and (2) w.r.t. x ; we have

Question 53.
Derivative of x² w.r.t x³ is ……………
Solution:
Let y = x²
and z = x³
We want to find \(\frac { dy }{ dz }\)
Diff. eqns. (1) and (2) w.r.t. x ; we have