d/(dθ)(5/(9sin(θ)+7cos(θ)))

解答

\frac{d}{d θ} (\frac{5}{9 sin ( θ ) + 7 cos ( θ )})

- \frac{5 ( 9 cos ( θ ) - 7 sin ( θ ) )}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}}

求解步骤

\frac{d}{d θ} (\frac{5}{9 sin ( θ ) + 7 cos ( θ )})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d θ} (\frac{1}{9 sin ( θ ) + 7 cos ( θ )})

使用指数法则:

\frac{1}{a} = a^{- 1}

= 5 \frac{d}{d θ} ((9 sin (θ) + 7 cos (θ))^{- 1})

使用连锁律:

- \frac{1}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}} \frac{d}{d θ} (9 sin (θ) + 7 cos (θ))

= - \frac{1}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}} \frac{d}{d θ} (9 sin (θ) + 7 cos (θ))

\frac{d}{d θ} (9 sin (θ) + 7 cos (θ)) = 9 cos (θ) - 7 sin (θ)

= 5 (- \frac{1}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}} (9 cos (θ) - 7 sin (θ)))

化简

5 (- \frac{1}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}} (9 cos (θ) - 7 sin (θ))) : - \frac{5 ( 9 cos ( θ ) - 7 sin ( θ ) )}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}}

= - \frac{5 ( 9 cos ( θ ) - 7 sin ( θ ) )}{( 9 sin ( θ ) + 7 cos ( θ ) ) ^{2}}

\frac{d}{d x} (3 (e^{x} + x e^{x}))

\int - 9 tan^{3} (x) d x

\int_{0}^{π} sec^{2} (\frac{π}{3}) d x

\int \frac{( x + 1 )}{x ^{2} + 7 x + 10} d x

\frac{d r}{d θ} = (r + e^{- θ}) tan (θ)