Question

nth derivative of sin^3x cos^5x dx​

Answer 1

Step-by-step explanation:

Let y=sin3xcos5x ……..(1)

we have to find

dx

2

d

2

y

=?

on differentiating equation (1) wrt x we get

dx

dy

dx

d

(sin3xcos5x)

dx

dy

=3cos3xcos5x−5sin5xsin3x [using multiplication rule]

again differentiating wrt x we get

dx

2

d

2

y

=3

dx

d

(cos3xcos5x)−5

dx

d

(sin5xsin3x)

=3[−3sin3xcos5x−5cos3xsin5x]−5[5cos5xsin3x+3cos3xsin5x]

⇒−9sin3xcos5x+15cos3xsin5x−25cos5xsin3x−15cos3xsin5x

⇒−34sin3xcos5x.