Question

find the nth derivative of x.sin x​

Answer 1

Step-by-step explanation:

Starting with the given function:

f

(

0

)

(

x

)

=

x

sin

x

Using the product rule we compute the first derivative:

f

(

1

)

(

x

)

=

x

(

d

d

x

sin

x

)

+

(

d

d

x

x

)

sin

x

=

x

cos

x

+

sin

x

Similarity, for the second derivative

f

(

2

)

(

x

)

=

x

(

d

d

x

cos

x

)

+

(

d

d

x

x

)

sin

x

+

d

d

x

sin

x

=

−

x

sin

x

+

cos

x

+

cos

x

=

2

cos

x

−

x

sin

x

And further derivatives:

f

(

3

)

(

x

)

=

−

2

sin

x

−

(

x

cos

x

+

sin

x

)

=

−

3

sin

x

−

x

cos

x

f

(

4

)

(

x

)

=

−

3

cos

x

−

(

−

x

sin

x

+

cos

x

)

=

−

4

cos

x

+

x

sin

x

So, By exploiting the phase shift properties, we have:

y

(

n

)

=

n

sin

(

x

+

(

n

−

1

)

π

2

)

+

x

sin

(

x

+

n

π

2

)

a

Answer 2

given

$f(x) = xsinx \\ \\ {f}^{(0)} (x) = xsinx$

Using product rule we compute first derivative

$▪\: {f}^{(1)} (x) = sinx + xcosx \\ \\ = 1.sin(x + \frac{(1 - 1)\pi}{2} ) + xsin(x + 1. \frac{\pi}{2} )$ similarly $\\ \\ ▪\: {f}^{(2)} (x) = cosx - xsinx + cosx \\ \\ = 2cosx - xsinx \\ \\ = 2.sin( x + \frac{(2 - 1)\pi}{2} ) + xsin(x + 2 \frac{\pi}{2} )\\ \\ \\ ▪ \: {f}^{(3)} (x) = - 2sinx - xcosx - sinx \\ \\ = - 3sinx - xcosx \\ \\ = 3sin(x + \frac{(3 - 1)\pi}{2}) + xsin(x + 3. \frac{\pi}{2} )\\ \\ \\▪\: {f}^{(4)} (x) = - 3cosx + xsinx - cosx \\ \\ = - 4cosx + xsinx \\ \\ = 4sin(x + \frac{(4 - 1)\pi}{2} ) + xsin(x + 4. \frac{\pi}{2}) \\ \\ . \: \: . \: \: . \: \: . \: \: . \: \: . \: \: . \: \: . \: \: .$

proceeding in this manner we get nth derivative as

${f}^{(n)} (x) = nsin(x + \frac{(n - 1)\pi}{2} ) + xsin(x + n. \frac{\pi}{2} )$