[tex] \begin{aligned} f(x) &= 3(4x^2-9)^2 \\ &\downarrow \\ f'(x) &= 6(4x^2-9)(8x) \\f'(x) &= 48x (4x^2-9) \\ f'(x) &= 192x^3-432x\end{aligned} [/tex]
[tex] \begin{aligned} f(x) &= 8 \sqrt {3x^2-3} \\ f(x) &= 8(3x^2-3)^{\frac{1}{2}} \\ &\downarrow \\ f'(x) &= 4(3x^2-3)^{-\frac{1}{2}}(6x) \\f'(x) &= \frac{24x }{(3x^2-3)^{\frac{1}{2}} }\\ f'(x) &= \frac{ 24x}{ \sqrt{3x^2-3}}\end{aligned} [/tex]
[tex]\begin{aligned} f(x) &= x^3-3x^2-24x-7 \\&\downarrow \\ f'(x) &= 3x^2-6x-24 \end{aligned} [/tex]
[tex] \begin{aligned} f'(x) &= 0 \\ 3x^2-6x-24 &= 0 \\ x^2-2x-8 &= 0 \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x&= 4\end{aligned} [/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
Soal 1
[tex] \begin{aligned} f(x) &= 3(4x^2-9)^2 \\ &\downarrow \\ f'(x) &= 6(4x^2-9)(8x) \\f'(x) &= 48x (4x^2-9) \\ f'(x) &= 192x^3-432x\end{aligned} [/tex]
Soal 2
[tex] \begin{aligned} f(x) &= 8 \sqrt {3x^2-3} \\ f(x) &= 8(3x^2-3)^{\frac{1}{2}} \\ &\downarrow \\ f'(x) &= 4(3x^2-3)^{-\frac{1}{2}}(6x) \\f'(x) &= \frac{24x }{(3x^2-3)^{\frac{1}{2}} }\\ f'(x) &= \frac{ 24x}{ \sqrt{3x^2-3}}\end{aligned} [/tex]
Soal 3
[tex]\begin{aligned} f(x) &= x^3-3x^2-24x-7 \\&\downarrow \\ f'(x) &= 3x^2-6x-24 \end{aligned} [/tex]
[tex] \begin{aligned} f'(x) &= 0 \\ 3x^2-6x-24 &= 0 \\ x^2-2x-8 &= 0 \\ (x+2)(x-4) &= 0 \\ x = -2 \text{ atau } x&= 4\end{aligned} [/tex]