Derivative of Polynomial functions

We know that df/dx = dg/dx + dh/dx when f(x) = g(x) + h(x)
We can extend it to any number of functions f1 = g1 + g2 + g3 ...
then df1/dx = dg1/dx + dg2/dx + dg3/dx + ...

If the polynomial function is f(x) = a0xn + a1xn-1 + a2xn-2 + a3xn-3 + a4xn-4 + ... then
df(x)/dx = n a0xn-1 + (n-1)a1xn-2 + (n-2)a2xn-3 + (n-3)a3xn-4 + ...

Problem: If the polynomial function is f(x) = 8x4 + 5x2 + 3x1 + 8 then find f '(x). Here  f '(x) = df/dx.
Solution:
As we know that df1/dx = dg1/dx + dg2/dx + dg3/dx + ...
df(x)/dx = d(8x4)/dx + d(5x2)dx + d(3x1) + d(8)/dx
df(x)/dx = 4×8x4-1 + 2×5x2-1 + 1×31-1 + 0.8x0-1
df(x)/dx = 32x3 + 10x + 3



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