These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-5-ex-5-5/
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5
Ex 5.5 Class 12 NCERT Solutions Question 1.
cos x. cos 2x. cos 3x
Solution:
Let y = cos x. cos 2x . cos 3x,
Taking log on both sides,
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x,
Differentiating w.r.t. x, we get
Exercise 5.5 Class 12 Maths Solutions Question 2.
\(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)
Solution:
Ex 5.5 Class 12 Maths Ncert Question 3.
(log x)cosx
Solution:
let y = (log x)cosx
Taking log on both sides,
log y = log (log x)cosx
log y = cos x log (log x),
Differentiating w.r.t. x,
Exercise 5.5 Class 12 NCERT Solutions Question 4.
x – 2sinx
Solution:
let y = x – 2sinx,
∴ y = u – v
Differentiating w.r.t. x,
Ch 5 Maths Class 12 Ex 5.5 NCERT Solutions 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 on both side,
logy = log [(x + 3)2 . (x + 4)3 . (x + 5)4]
= log (x + 3)2 + log (x + 4)3 + log (x + 5)4
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get
On simplification,
\(\frac { dy }{ dx }\) = (x + 3)(x + 4)²(x + 5)³(9x² + 70x + 133)
Question 6.
\({ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)
Solution:
Let u = \(\left(x+\frac{1}{x}\right)^{x}\) = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)
v = \(x^{\left(1+\frac{1}{x}\right)}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)
Taking logaritham on both sides,
logu = x log \(\left(\frac{x^{2}+1}{x}\right)\)
∴ log u = x(log(x²+ 1) – logx)
Differentiating both sides w.r.t.x,
v = \(x^{\left(1+\frac{1}{x}\right)}\)
Taking logaritham on both sides,
logv = (1 + \(\frac { 1 }{ x }\)) log x
Differentiating both sides w.r.t. x,
Question 7.
(log x)x + xlogx
Solution:
Let u = (log x)x, v = xlogx and y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = (log x)x
Taking logaritham on both sides,
logu = x log(log x)
Differentiating both sides w.r.t. x,
v = xlog x
Taking logarithm on both sides,
logv = logx. logx = (logx)²
Differentiating both sides w.r.t. x,
Question 8.
(sin x)x+sin-1 \(\sqrt{x}\)
Solution:
Let y = (sin x)x + sin-1 \(\sqrt{x}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = (sin x)x
\(\frac { du }{ dx }\) = (sin x)x[x cotx + log sinx] … (2)
Let y = (sinx)x
Taking logarithm on bath sides,
logy = x.log sin x
Differentiating both sides w.r.t. x,
Question 9.
xsinx + (sin x)cosx
Solution:
Let y = xsinx + (sin x)cosx
Differentiating both sides w.r.t. x,
(i) Let y = (sin x)x
Taking logarithm on both sides,
logy = sinx. logx
Differentiating both sides w.r.t. x,
(ii) Let y = (sinx)cosx
Taking logarithm on both sides,
logy = cosx. log sinx
Differentiating both sides w.r.t. x,
Question 10.
\({ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \)
Solution:
Let u = xx cosx and v = \(\frac{x^{2}+1}{x^{2}-1}\)
Let y = u + v
Differentiating w.r.t. x,
∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)
u = xx cosx
Taking logarithm on both sides,
logu = x cosx . logx
Differentiating both sides w.r.t. x,
Differentiating both sides w.r.t. x,
Question 11.
\((x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}\)
Solution:
Let u = (x cosx)x and v = \((x \sin x)^{\frac{1}{x}}\)
u = (x cosx)x
Taking logarithm on both sides,
log u = x log (x cosx)
log u = x[log x + log cosx]
Differentiating w.r.t. x,
Question 12.
xy + yx = 1
Solution:
xy + yx = 1
let u = xy and v = yx
∴ u + v = 1,
\(\frac { du }{ dx } +\frac { dv }{ dx }=0\)
Now u = xy
Taking logarithm on both sides,
log u = y log x
Differentiating w.r.t. x
Question 13.
yx = xy
Solution:
xy = yx
Taking logarithm on both sides,
y log x = x log y
Differentiating w.r.t. x
Question 14.
(cos x)y = (cos y)x
Solution:
(cos x)y = (cos y)x
Taking logarithm on both sides,
y log cosx = x log cosy
Differentiating both sides w.r.t. x,
Question 15.
xy = e(x-y)
Solution:
log(xy) = log e(x-y)
⇒ log(xy) = x – y
⇒ logx + logy = x – y
\( ⇒\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } ⇒\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)
Question 16.
Find the derivative of the function given by f (x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) and hence find f'(1).
Solution:
Let f(x) = y = (1 + x)(1 + x2)(1 + x4)(1 + x8)
Taking log both sides, we get
logy = log [(1 + x)(1 + x2)(1 + x4)(1 + x8)]
logy = log(1 + x) + log (1 + x2) + log(1 + x4) + log(1 + x8)
Differentiating w.r.t. x,
Question 17.
Differentiate (x2 – 5x + 8) (x3 + 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:
(i) By using product rule
f’ = (x2 – 5x + 8) (3x2 + 7) + (x3 + 7x + 9) (2x – 5)
f = 5x4 – 20x3 + 45x2 – 52x + 11.
(ii) y = x5 – 5x4 + 15x³ – 26x² + 11x + 72
(by expanding the product)
Differentiating w.r.t. x, dy
\(\frac { dy }{ dx }\) = 5x4 – 20x³ + 45x² – 52x + 11
(iii) y = (x² – 5x + 8)(x³ + 7x + 9)
Taking logarithm on both sides,
log y = log(x² – 5x 4- 8)(x³ + 7x + 9)
log y = log(x² – 5x + 8) + log(x³ + 7x + 9)
Differentiating, w.r.t x,
= (2x – 5) (x³ + 7x + 9) + (3x² + 7) (x² – 5x + 8)
= 5x4 – 20x³ + 45x² – 52x + 11
Yes, the answers are the same.
Question 18.
If u, v and w are functions of w 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, second by logarithmic differentiation.
Solution:
Let y = u.v.w
(a) Product rule
(b) Logarithmic differentiation
y = uvw
Taking logarithm on both sides, we get
log y = log uvw
i.e., log y = log u + log v + log w
Differentiating both sides w.r.t. x, we get
