Question

Tentukan turunan dari
[tex]\rm (2x+1)^3(x^2-5x+1)^2[/tex]​

Answer 1

Penjelasan dengan langkah-langkah:

FungSi

f(x) = (2x+1)³(x²-5x+1)²

...... p = (2x+1)³ ....... q = (x²-5x+1)²

...... u' = 2x+1 → 2 ........ r' = x²-5x → 2x-5

....... p' = u'.d/dx (2x+1)³

= 2.3(2x+1)²

= 6(2x+1)²

= 6(4x²+4x+1)

= 24x²+24x+6

........q' = r'.d/dx (x²-5x+1)²

= (2x-5)(2(x²-5x+1)

= (2x-5)(2x²-10x+2)

= 4x³-20x²+4x-10x²+50x-10

= 4x³-30x²+54x-10

f'(x) = p'q + q'p

f'(x) = (24x²+24x+6)(x²-5x+1)² + (4x³-30x²+54x-10)(2x+1)³

Answer 2

Jawab:

[tex]\displaystyle 2(x+1)^2(x^2-5x+1)(7x^2-23x-2)[/tex]

Penjelasan dengan langkah-langkah:

Turunan perkalian dua fungsi [tex]\displaystyle \frac{d}{dx}(uv)=u'v+uv'[/tex]

Turunan fungsi berpangkat [tex]\displaystyle y=[f(x)]^n\rightarrow y'=n[f(x)]^{n-1}f'(x)[/tex]

[tex]\displaystyle u=(2x+1)^3\rightarrow u'=6(2x+1)^2\\v=(x^2-5x+1)^2\rightarrow v'=2(x^2-5x+1)(2x-5)[/tex]

Diperoleh

[tex]\begin{aligned}\frac{d}{dx}(2x+1)^3(x^2-5x+1)^2&\:=6(x^2-5x+1)^2(2x+1)^2+2(2x-5)(x^2-5x+1)(2x+1)^3\\\:&=(x^2-5x+1)(2x+1)^2~[6(x^2-5x+1)+2(2x-5)(2x+1)]\\\:&=(x^2-5x+1)(2x+1)^2~[6x^2-30x+6+2(4x^2-8x-5)]\\\:&=(x^2-5x+1)(2x+1)^2(6x^2-30x+6+8x^2-16x-10)\\\:&=(x^2-5x+1)(2x+1)^2(14x^2-46x-4)\\\:&=2(x^2-5x+1)(7x^2-23x-2)(2x+1)^2\end{aligned}[/tex]