Question

find the derivative of xsinx/(2x+3)​

Answer 1

Answer:

Step-by-step explanation:

d

y

d

x

=

x

cos

x

+

sin

x

Explanation:

We have:

y

=

x

sin

x

Which is the product of two functions, and so we apply the Product Rule for Differentiation:

d

d

x

(

u

v

)

=

u

d

v

d

x

+

d

u

d

x

v

, or,  

(

u

v

)

'

=

(

d

u

)

v

+

u

(

d

v

)

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

So with  

y

=

x

sin

x

;

{

Let

u

=

x

⇒

d

u

d

x

=

1

And

v

=

sin

x

⇒

d

v

d

x

=

cos

x

Then:

d

d

x

(

u

v

)

=

u

d

v

d

x

+

d

u

d

x

v

Gives us:

d

d

x

(

x

sin

x

)

=

(

x

)

(

cos

x

)

+

(

1

)

(

sin

x

)

 

∴

d

y

d

x

=

x

cos

x

+

sin

x

Answer 2

Answer:

$\dfrac{d}{dx} ( \dfrac{x \: sinx}{2x + 3} ) \:$

Use the formula ,

$\dfrac{d}{dx} ( \dfrac{f}{g} ) = \dfrac{g {f}^{l} - f {g}^{l} }{ {g}^{2} }$

» (2x+3 d/dx (x sinx) - xsinx d/dx (2x +3))/(2x+3)^2

$\dfrac{(2x + 3) \: ( cosx) - (x \: sinx)(2)}{(2x + 3)(2x + 3)}$

$\implies \: \dfrac{2x \: cosx + 3cosx - 2xsinx}{4 {x}^{2} + 12x + 9}$