derivative of 10^{tan(2x})

解答

\frac{d}{d x} (1 0^{t a n (2 x)})

ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)

求解步骤

\frac{d}{d x} (1 0^{t a n (2 x)})

使用指数法则:

a^{b} = e^{b l n (a)}

1 0^{t a n (2 x)} = e^{t a n (2 x) l n (10)}

= \frac{d}{d x} (e^{t a n (2 x) l n (10)})

使用连锁律:

e^{t a n (2 x) l n (10)} \frac{d}{d x} (tan (2 x) ln (10))

= e^{t a n (2 x) l n (10)} \frac{d}{d x} (tan (2 x) ln (10))

\frac{d}{d x} (tan (2 x) ln (10)) = ln (10) sec^{2} (2 x) \cdot 2

= e^{t a n (2 x) l n (10)} ln (10) sec^{2} (2 x) \cdot 2

化简

e^{t a n (2 x) l n (10)} ln (10) sec^{2} (2 x) \cdot 2 : ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)

= ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)

y^{^{''}} + 16 y = sin (4 x)

\int \frac{cosh ( x )}{1 + sinh ( x )} d x

\int \frac{1}{( e ^{x} + 1 )} d x

\frac{d}{d x} (\frac{1 + sin ( x )}{cos ( x )})

y^{^{'}} - \frac{4 y}{x} = - \frac{2}{x ^{2}}