Réitigh tan(4x)

$$\tan ( 4x )$$

Difreálaigh w.r.t. x

$$\frac{4\left(16\left(\sin(x)\right)^{2}\left(\cos(x)\right)^{6}+16\left(\cos(x)\right)^{2}\left(\sin(x)\right)^{6}+2\left(\sin(2x)\right)^{4}-4\left(\sin(2x)\right)^{2}+1\right)}{4\left(\sin(2x)\right)^{4}-4\left(\sin(2x)\right)^{2}+1}$$

Céimeanna ag Baint Úsáid as an Riail Slabhra

Más F comhshuíomh dhá fheidhm indifreáilte f\left(u\right) agus u=g\left(x\right), is é sin, más F\left(x\right)=f\left(g\left(x\right)\right), mar sin is ionann díorthach F agus díorthach f maidir le u méadaithe faoi dhíorthach g maidir le x, is é sin, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

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}.

$$\left(\sec(4x^{1})\right)^{2}\times 4x^{1-1}$$

Simpligh.

$$4\left(\sec(4x^{1})\right)^{2}$$

Do théarma ar bith t, t^{1}=t.

$$4\left(\sec(4x)\right)^{2}$$

Luacháil

$$\tan(4x)$$