Find the derivative of the following functions:f (x)= x^8 sin 5x

Expert Answers
changchengliang eNotes educator| Certified Educator

Use product rule:

f'(x) = U.dV/dx + V.dU/dx

Let U = x^8  and V = sin 5x


= x^8 . [d(sin 5x)/d(5x) . d(5x)/dx] + (sin 5x) . 8x^7

= x^8 . cos 5x . (5)  +  8 . (x^7) . sin 5x

= 5.(x^8).cos 5x  +  8.(x^7).sin 5x


Just a remark:
Notice the power of chain rule as we try to differentiate seemingly more complicated expressions.  I always liked doing differentiation over integration for this simple reason --  that in differentiation, because of the existence of chain rule, one can never come to a deadlock.

hala718 eNotes educator| Certified Educator

f (x)= x^8 sin 5x

We need to find the first derivative:

Let f(x) = u*v such that:

u= x^8   ==>   u' = 8x^7

v= sin5x   ==>   v' = 5cos5x

Then we know that:

f'(x) = ( u'v + uv')

         = ( x^8* 5cos5x) + 8x^7 * sin5x

          = 5cos5x*x^8 + 8sin5x *x^7

We will factor x^7 from both sides:

==> f'(x) = x^7 ( 5x*cos5x + 8sin5x)

neela | Student

f(x) = x^8 sinx.

To differentiate x^8sin5x, which is the product of x^8 and sin5x.

We use the product formula  to differentiate the function.

(u(x)v(x))' = u'(x)v(x)+u(x)v'(x).

Also  we use (f(g(x))' = (df/dg)*g'(x) to differentiate a function of function.

Therefore (x^8*sin5x}' =  (x^8)'sin5x +x^8 (sin5x)'

(x^8*sin5x}' = 8x^7 sin5x + x^8 (cos5x) (5x)'

(x^8*sin5x}' = 8x^7 sin5x +x^8 (cos5x) 5

(x^8*sin5x}' = 8x^7 x sin5x +5x^8 cos5x.

(x^8*sin5x}' = x^7 (8sin5x+5xcos5x).

giorgiana1976 | Student

To calculate the first derivative of the given function, we'll use the product rule and the chain rule:

f'(x) = (x^8)*(sin 5x)

We'll have 2 functions g and h:

(g*h)' = g'*h + g*h'

We'll put g = x^8 => g' = 8x^7

We'll put h = sin 5x => h' = (cos 5x)*(5x)'

h' = 5cos 5x

We'll substitute g,h,g',h' in the expression of (g*h)':

(g*h)' = 8x^7*(sin 5x) + x^8*(5cos 5x)

We'll factorize by x^7:

f'(x) = x^7[8(sin 5x) + x*(5cos 5x)]