TS EAMCET 4-Aug-2021 Shift 2 Question Paper

(a+bx)ey∕x=x
Taking log on both the sides, we get
log(a+bx)+‌

y

x

=log‌x
or y=x[log‌x−log(a+bx)].....(i)
Differentiating w.r.t. x,
y′=x(‌

1

x

−‌

b

a+bx

)+[log‌x−log(a+bx)]
y′=1−‌

xb

a+bx

−‌

y

x

...(ii)
or ‌

b

a+bx

=‌

1

x

+‌

y

x2

−‌

y′

x

....(iii)
Again, differentiating Eq. (ii) w.r.t. x,
y′′‌‌=−b[‌

(a+bx)−bx

(a+bx)2

]+‌

y′

x

−‌

y

x2

‌‌=‌

xb2

(a+bx)2

−‌

b

(a+bx)

+‌

y′

x

−‌

y

x2

From Eq. (iii),
=‌‌‌

y2

x3

+‌

y′2

x

−‌

2yy′

x2

=‌‌‌

1

x3

[y2+(xy1)2−2xyy′]
=‌‌‌

1

x3

(xy′−y)2