How do you use the Product Rule to find the derivative of (x^2)(x^3+4)?

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The description is attached underneath.

for differentiation where result of functions is attached.

If ##f(x)## and ##g(x)## are two functions then their result would be ##f(x)*g(x)##

The derivative ##(f(x)*g(x))'## is attached by the result rule

##(f(x)g(x))' = f(x)g'(x)+g(x)f'(x)##

Derivative rules

##d/dx (x^n) = nx^(n-1)## ##d/dx(f(x)+g(x)) = d/dx(f(x)) + d/dx(g(x))## ##d/dx(c) = 0## Derivative of fixed is naught.

Now future to our problem

##y=(x^2)(x^3+4)##

##f(x) = x^2## and ##g(x) = (x^3+4)##

##f'(x) = d/dx(x^2) => f'(x) = 2x## ##g'(x) = d/dx(x^3+4) = d/dx (x^3) + d/dx(4)## ##g'(x) = 3x^2 + 0## ##g'(x) = 3x^2##

Using result rule

##d/dx((x^2)(x^3+4)) = x^2d/dx(x^3+4) + (x^3+4)d/dx(x^2)##

##= x^2(3x^2) + (x^3+4)(2x)##

##= 3x^4 + 2x^4+8x## simplifying

##=5x^4+8x## Answer

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