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