# GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Question 1.
cos x . cos 2x . cos 3x
Solution:
Let y – cos x . cos 2.x . cos 3x.
Taking log of both sides, we get
log y = log (cos x . cos 2x . cos 3x)
= log cos x + log cos 2x + log cos 3x [Note : loga mn = logam + loga n]
Differentiating both sides w.r.t. x,

Question 2.
$$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Solution:
Let y = $$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$$
Taking log of both sides, we get

Question 3.
$$(\log x)^{\cos x}$$
Solution:
Let y = $$(\log x)^{\cos x}$$
Taking log of both sides, we get
log y = log (log x)cos x
= cos x log (log x) [log mn = n log m]
Differentiating both sides w.r.t. x,

Question 4.
xx – 2sin x
Solution:
Let y = xx – 2sin x
= u – v.
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ – $$\frac { dv }{ dx }$$.
Now, u = x² or log u = log xsin x = x log x [ ∵ log msin n = n log m]
Differentiating w.r.t. x, we get
$$\frac { 1 }{ u }$$ $$\frac { du }{ dx }$$ = 1 . log x + x . $$\frac { 1 }{ x }$$ = (1 + log x)
∴ $$\frac { du }{ dx }$$ = u (1 + log x) = xsin x(1 + log x).
Taking, v = 2sin x, we get
∴ log v = log 2sin x = sin x log 2
∴ $$\frac { 1 }{ v }$$ $$\frac { dv }{ dx }$$ = cos x log 2.
∴ $$\frac { dv }{ dx }$$ = v cos x log 2 = 2sin x cos x log 2.
∴ $$\frac { dv }{ dx }$$ = $$\frac { du }{ dx }$$ – $$\frac { dv }{ dx }$$ = xx (1 + log x) – 2sin x cos x log 2.

Question 5.
(x + 3)² . (x + 4)³. (x + 5)4
Solution:
Let y = (x + 3)². (x + 4)³. (x + 5)4 .
Taking log of both sides, we get .
log y = log [(x + 3)².(x + 4)³.(x + 5)4]
= log (x + 3)² + log (x + 4)³ + log (x + 5)4 [log mn = log m + log n]
= 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5) [log mn = n log m]
Differentiating both sides w.r.t. x,
$$\frac { 1 }{ y }$$ $$\frac { dy }{ dx }$$ = $$\frac { 2 }{ x+3 }$$ + $$\frac { 3 }{ x+4 }$$ + $$\frac { 4 }{ x+5 }$$
∴ $$\frac { dy }{ dx }$$ = y($$\frac { 2 }{ x+3 }$$ + $$\frac { 3 }{ x+4 }$$ + $$\frac { 4 }{ x+5 }$$)
= (x + 3)² . (x + 4)³. (x + 5)4($$\frac { 2 }{ x+3 }$$ + $$\frac { 3 }{ x+4 }$$ + $$\frac { 4 }{ x+5 }$$)

Question 6.
$$\left(x+\frac{1}{x}\right)^{x}$$ + $$x^{\left(1+\frac{1}{x}\right)}$$
Solution:
$$\left(x+\frac{1}{x}\right)^{x}$$ + $$x^{\left(1+\frac{1}{x}\right)}$$
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = (x + $$\frac { 1 }{ x }$$)x .
Taking log both sides, we get

Now v = $$x^{\left(1+\frac{1}{x}\right)}$$.
Taking log of both sides, we get
log v = log $$x^{\left(1+\frac{1}{x}\right)}$$
= (1 + $$\frac { 1 }{ x }$$) log x [log mn = n log m]
Differentiating w.r.t. x,

Question 7.
(logx)x + xlog x
Solution:
Let y = (logx)x + xlog x
= u + v
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = (log x)x.
Taking log both sides, we get
log v = log (log x)x
= x log (log x) [log mn = n log m]
Differentiating w.r.t. x,

Now, v = xlog x.
Taking log of both sides, we get
log v = log xlog x [log mn = n log m]
= log x . log x = (log x)²
Differentiating w.r.t. x,

Question 8.
(sin x)x + sin-1$$\sqrt{x}$$
Solution:
Let y = (sin x)x + sin-1$$\sqrt{x}$$
= u + v
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = (sin x)x.
Taking log both sides, we get
log u = log (sin x)x
= x log (sin x) [log mn = n log m]
Differentiating w.r.t. x,

Question 9.
xsin x + (sin x)cos x
Solution:
Let y = xsin x + (sin x)cos x
= u + v
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = xx.
Taking log both sides, we get
log u = log xx [log mn = n log m]
= sin x log x
Differentiating w.r.t. x,

Now, u = (sin x)cos x.
Taking log both sides, we get
log v = log (sin x)cos x [log mn = n log m]
Differentiating w.r.t. x,

Question 10.
xx cos x + $$\frac{x^{2}+1}{x^{2}-1}$$
Solution:
Let y = xx cos x + $$\frac{x^{2}+1}{x^{2}-1}$$
= u + v
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = xx cos x.
Taking log both sides, we get
log u = log xx cos x
= x cos x log x [log mn = n log m]
Differentiating w.r.t. x,

Question 11.
(x cos x)x + (x sin x)$$\frac{1} {x}$$
Solution:
Let y = (x cos x)x + (x sin x)[$$\frac{1} {x}$$
= u + v
∴ $$\frac { dy }{ dx }$$ = $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$.
Now, u = (x cos x)x.
Taking log both sides, we get
log u = log (x cos x)x
= x log (x cos x) [log mn = n log m]
= x (log x + log cos x)
Differentiating w.r.t. x,

Further, v = (x sinx)$$\frac{1} {x}$$.
Taking log of both sides, we get
log v = log (x sinx)$$\frac{1} {x}$$
= $$\frac{1} {x}$$ log (x sin x) [log mn = n log m]
= $$\frac{1} {x}$$ (log x + log sin x)
Differentiating both sides,

Question 12.
xy + yx = 1
Solution:
Let y = xy + yx = 1
Let u = xy and u = yx
∴ u + v = 1
or $$\frac { du }{ dx }$$ + $$\frac { dv }{ dx }$$ = 0. … (1)
Now, u = xy.
Taking log both sides, we get
log u = log xy
= y log x [log mn = n log m]
Differentiating w.r.t. x,

Question 13.
yx = xy
Solution:
yx = xy
Taking log both sides, we get
x log y = y log x [log mn = n log m]
Differentiating w.r.t. x,

Question 14.
(cos x)y = (cos y)x
Solution:
(cos x)y = (cos y)x
Taking log both sides, we get
y log (cos x) = x log cos y [log mn = n log m]
Differentiating w.r.t. x,

Question 15.
xy = ex-y
Solution:
xy = ex-y
Taking log both sides, we get
log x + log y = (x – y) log e [log xy = log x + log y]
= (x – y) [ loge = 1]
Differentiating w.r.t. x,
$$\frac { 1 }{ x }$$ + $$\frac { 1 }{ y }$$$$\frac { dy }{ dx }$$ = 1 – $$\frac { dy }{ dx }$$
⇒ ($$\frac { 1 }{ y }$$ + 1)$$\frac { dy }{ dx }$$ = 1 – $$\frac { 1 }{ x }$$ = $$\frac { x-1 }{ x }$$
∴ $$\frac { dy }{ dx }$$ = $$\frac { y(x-1) }{ x(y+1) }$$

Question 16.
Find the derivatives of the functions given by
f(x) = (1 + x) (1 + x²) (1 + x4) (1 + x8) and hence find f’ (1).
Solution:
Let y = (1 + x) (1 + x²) (1 + x4) (1 + x8)
Taking log of both sides, we get
log y = log(1 + x) (1 + x²) + log(1 + x2) + log(1 + x4) + log(1 + x8)
Differentiating w.r.t. x,

Question 17.
Differentiate (x² – 5x + 8)(x² + 7x + 9) in three ways mentioned below?
(i) by using product rule.
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Solution:
Method (i) : (uv)’ = u’v + uv’
Let y = (x² -5x + 8)(x³ + 7x + 9)
Differentiating by product rule, we get
$$\frac { dy }{ dx }$$ = [ $$\frac { d }{ dx }$$(x² -5x + 8) ] (x³ + 7x + 9) + (x² – 5x + 8) x $$\frac { d }{ dx }$$ (x³ + 7x + 9)
= (2x – 5) (x³ + 7x + 9) + (x² – 5x + 8) (3x² + 7)
= 2x(x³ + 7x + 9) – 5(x³ + 7x + 9) + 3x²(x² – 5x + 8) + 7(x² – 5x + 8)
= 5x4 – 20x³ + 45x² – 52x + 11.

Method (ii) : y = (x² – 5x + 8) (x³ + 7x + 9)
= x² (x³ + 7x + 9) – 5x (x³ + 7x + 9) + 8 (x³ + 7x + 9)
= x5 – 5x4 + 15x³ – 26x² + 11 x + 72
$$\frac { dy }{ dx }$$ = 5x4 – 20x³ + 45x² – 52 x + 11.

Method (iii) : y = (x² – 5x + 8) (x³ + 7x + 9)
Taking log of both sides, we get
log y = log (x² – 5x + 8) + log (x³ + 7x + 9) [log mn = log m + log n]
Differentiating w.r.t. x,

[See part (ii) for simplification] Obviously, the answer is the same in all the three cases.

Question 18.
If u, v and w are functions of x, then show that $$\frac { d }{ dx }$$(u . v . w) = $$\frac { du }{ dx }$$ . v . w + u . $$\frac { dv }{ dx }$$ . w + u . v . $$\frac { dw }{ dx }$$ in two ways : first by repeated application of product rule and secondly by logarithmic differentiation.
Solution:
Let y = u .v . w = u . (vw)
(i) Differentiating both sides we get