How do you use the product rule to differentiate #f(x)=(6x+5)(x^3-2)#?

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Answer 1 · 2018-05-05T16:11:42

#f'(x) = 24x^3 + 3#

This is actually a bit easy than the other ones, so make sure to understand this.

So, The Product Rule of Differentiation is:-

If #h(x) = f(x) * g(x)#, then, #h'(x) = f'(x) * g(x) + f(x) * g'(x)# .

So, Using the Rule,

#f'(x) = d/dx(6x + 5) * (x^3 - 2) + (6x + 5) * d/dx(x^3 -2)#

#= {d/dx (6x) + d/dx(5)} * (x^3 - 2) + (6x + 5) * {d/dx(x^3) - d/dx(2)}#

#= (6 + 0) * (x^3 - 2) + (6x + 5) * (3* x^(3 -1) - 0)#

[Using the identities :-
#d/dx("constant") = 0#, #d/dx(mx) = m# where #m# is a constant and #d/dx(x^n) = nx^(n - 1)#
]

#= 6(x^3 - 2) + 3x^2(6x + 5) = 6x^3 - 12 + 18x^3 + 15 = 24x^3 + 3#

So, #f'(x) = 24x^3 + 3#.

Hope this helps.