Réitigh d/dxleft(x^4-x^3/3x+1right)

$$\frac{d}{d x } \left( \frac{ { x }^{ 4 } - { x }^{ 3 } }{ 3x+1 } \right)$$

Luacháil

$$\left(\frac{x}{3x+1}\right)^{2}\left(9x^{2}-2x-3\right)$$

Céimeanna ag Baint Úsáid as Riail na nDíorthach don Líon

Do dhá fheidhm indifreáilte ar bith, is ionann díorthach líon an dá fheidhme agus an t-ainmneoir méadaithe faoi dhíorthach an uimhreora lúide an t-uimhreoir méadaithe faoi dhíorthach an ainmneora, agus iad ar fad roinnte faoin ainmneoir cearnaithe.

$$\frac{\left(3x^{1}+1\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{4}-x^{3})-\left(x^{4}-x^{3}\right)\frac{\mathrm{d}}{\mathrm{d}x}(3x^{1}+1)}{\left(3x^{1}+1\right)^{2}}$$

Is ionann díorthach iltéarmaigh agus suim dhíorthaigh a théarmaí. Is ionann díorthach téarma thairisigh agus 0. Is ionann díorthach ax^{n} agus nax^{n-1}.

$$\frac{\left(3x^{1}+1\right)\left(4x^{4-1}+3\left(-1\right)x^{3-1}\right)-\left(x^{4}-x^{3}\right)\times 3x^{1-1}}{\left(3x^{1}+1\right)^{2}}$$

Simpligh.

$$\frac{\left(3x^{1}+1\right)\left(4x^{3}-3x^{2}\right)-\left(x^{4}-x^{3}\right)\times 3x^{0}}{\left(3x^{1}+1\right)^{2}}$$

Méadaigh 3x^{1}+1 faoi 4x^{3}-3x^{2}.

$$\frac{3x^{1}\times 4x^{3}+3x^{1}\left(-3\right)x^{2}+4x^{3}-3x^{2}-\left(x^{4}-x^{3}\right)\times 3x^{0}}{\left(3x^{1}+1\right)^{2}}$$

Méadaigh x^{4}-x^{3} faoi 3x^{0}.

$$\frac{3x^{1}\times 4x^{3}+3x^{1}\left(-3\right)x^{2}+4x^{3}-3x^{2}-\left(x^{4}\times 3x^{0}-x^{3}\times 3x^{0}\right)}{\left(3x^{1}+1\right)^{2}}$$

Chun cumhachtaí an bhoinn chéanna a mhéadú, suimigh a n-easpónaint.

$$\frac{3\times 4x^{1+3}+3\left(-3\right)x^{1+2}+4x^{3}-3x^{2}-\left(3x^{4}-3x^{3}\right)}{\left(3x^{1}+1\right)^{2}}$$

Simpligh.

$$\frac{12x^{4}-9x^{3}+4x^{3}-3x^{2}-\left(3x^{4}-3x^{3}\right)}{\left(3x^{1}+1\right)^{2}}$$

Cuir téarmaí cosúla le chéile.

$$\frac{9x^{4}-6x^{3}+4x^{3}-3x^{2}}{\left(3x^{1}+1\right)^{2}}$$

Do théarma ar bith t, t^{1}=t.

$$\frac{9x^{4}-6x^{3}+4x^{3}-3x^{2}}{\left(3x+1\right)^{2}}$$

Difreálaigh w.r.t. x

$$\frac{6x\left(9x^{3}+5x^{2}-x-1\right)}{\left(3x+1\right)^{3}}$$