Find The Derivative Or The Integral Of Y= -Sec 8x\Xb3

Find the derivative or the integral of y= -sec 8x³

 

y = -sec (8x^{3})

Derivative:

\frac{dy}{dx} = -sec (8x^{3})

u = 8x^{3}

du = 24x^{2}

\frac{d}{dx} sec xdx = secxtanxdx

-\frac{dy}{dx}sec (8x^{3})

-(24x^{2}sec(8x^{3})tan(8x^{3}))

y^{} = -24x^{2}sec(8x^{3})tan(8x^{3}))

Antiderivative:

∫ -sec (8x^{3}) dx

u = 8x^{3}

du = 24x^{2}

∫secxdx = ㏑lsecx + tanxl + C

-∫sec (8x^{3})dx

-(\frac{1}{24x^{2}} ㏑lsec(8x^{3}) + tan(8x^{3})l) + C

-\frac{1}{24x^{2}} ㏑lsec(8x^{3}) + tan(8x^{3})l + C


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