derivative of tan(pix-1+e^{cos(x)})

解答

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

sec^{2} (π x - 1) π - e^{c o s (x)} sin (x)

求解步骤

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (tan (π x - 1)) + \frac{d}{d x} (e^{c o s (x)})

\frac{d}{d x} (tan (π x - 1)) = sec^{2} (π x - 1) π

\frac{d}{d x} (e^{c o s (x)}) = - e^{c o s (x)} sin (x)

= sec^{2} (π x - 1) π - e^{c o s (x)} sin (x)

d er i v a t i v e y = (x - 1)^{3}

\int tan (x) ln (cos (x)) d x

\int \frac{x ^{2}}{x ^{2} + 4} d x

a re a y = x^{2} - 1, y = 2 x - 1, y = x + 1

t an g e n t y = \frac{2}{1 + x ^{2}}, (3, 0.2)