Answer: [tex]f'(x)=12x^2 - 6x+36[/tex]
Step-by-step explanation:
According to the product rule of derivatives.
[tex]f'(x)=u'(x)v(x)+u(x)v'(x)[/tex]
The given function : [tex]f(x) = (4x - 3)(x^2 + 9)[/tex]
here , [tex]u(x) = 4x - 3[/tex] and [tex]v(x)=x^2 + 9[/tex]
Differentiate both sides with respect to x, we get
[tex]u'(x) = 4+0=4[/tex] and [tex]v(x)=2x +0=2x[/tex]
[tex][\because \dfrac{d(ax)}{dx}=a\ , \dfrac{d(a)}{dx}=0 \ \&\ \ \dfrac{d(x^n)}{dx}=nx^{n-1}][/tex]
Then, Tge derivative of withe respect to x will be :
[tex]f'(x)=u'(x)v(x)+u(x)v'(x)[/tex]
[tex]=4(x^2 + 9)+( 4x - 3)(2x)[/tex]
[tex]=4x^2 +36+8x^2 - 6x[/tex]
[tex]=12x^2 - 6x+36 [/tex]
Hence, the derivative of the function is [tex]f'(x)=12x^2 - 6x+36[/tex] .
