Question

If y= x3/2 then find the value of dy/dx​

Answer 1

Answer:

x

(

x

2

−3)

.{

(

x

2

−3)

x

+(2x)logx}+

(x−3)

x

2

.{

x

2

(x−3)

+2xlog(x−3)}

x(x2-3).{(x2-3)x+(2x)logx}+(x-3)x2.{x2(x-3)+2xlog(x-3)}

Solution :

Let  

u=

x

(

x

2

−3)

andv=

(x−3)

x

2

.

u=x(x2-3)andv=(x-3)x2.

Then,  

y=u+v

y=u+v

Now,  

u=

x

(

x

2

−3)

u=x(x2-3)

⇒logu=(

x

2

−3)logx

⇒logu=(x2-3)logx

⇒

1

u

.

du

dx

=(

x

2

−3).

1

x

+(2x)logx

⇒1u.dudx=(x2-3).1x+(2x)logx

⇒

du

dx

=u.[

x

2

−3

x

+(2x)logx]=

x

(

x

2

−3)

.{

(

x

2

−3)

x

+(2x)logx}.

⇒dudx=u.[x2-3x+(2x)logx]=x(x2-3).{(x2-3)x+(2x)logx}.

And,  

v=

(x−3)

x

2

v=(x-3)x2

⇒logv=

x

2

log(x−3)

⇒logv=x2log(x-3)

⇒

1

v

.

dv

dx

=

x

2

.

1

(x−3)

+2xlog(x−3)

⇒1v.dvdx=x2.1(x-3)+2xlog(x-3)

⇒

dv

dx

=v.{

x

2

(x−3)

+2xlog(x−3)}=

(x−3)

x

2

{

x

2

(x−3)

+2xlog(x−3)}

Step-by-step explanation: