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

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### 4 Answers

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)**

Use product rule:

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

Let U = x^8 and V = sin 5x

f'(x)

= 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**

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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.

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).

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)]**