Respuesta :

Product rule:-

[tex]\boxed{\sf \dfrac{d}{dx}uv=uv'+vu'}[/tex]

Let's solve

[tex]\\ \sf\longmapsto \dfrac{d}{dx}(x^2+1)(x-1)[/tex]

[tex]\\ \sf\longmapsto (x^2+1)\dfrac{d}{dx}(x-1)+(x-1)\dfrac{d}{dx}(x^2+1)[/tex]

[tex]\\ \sf\longmapsto (x^2+1)+(x-1)(2x)[/tex]

[tex]\\ \sf\longmapsto x^2+1+2x^2-2x[/tex]

[tex]\\ \sf\longmapsto 3x^2-2x+1[/tex]

Answer:

  • Product rule:

[tex]\sf \frac{d}{dx}uv=uv'+vu'[/tex]

[tex]\longmapsto \sf ({x}^{2}+1)\times\: \frac{d}{dx}(x - 1)+\frac{d}{dx}(x- 1) × (x^2 +1)[/tex]

[tex] \longmapsto \sf ( {x}^{2}+1)+(x-1) (2x) [/tex]

[tex] \longmapsto \sf x^2+1+2x^2-2x [/tex]

[tex] \longmapsto \sf( {x}^{2} +1+ {2x}^{2} - 2x[/tex]

[tex]\longmapsto \sf 3x^2-2x+1[/tex]

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