d/(dθ)((A)(5pir(θ)^2-4r(θ)^2θ-4r(θ)^2sin(θ)))

פתרון

\frac{d}{d θ} ((A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ)))

A (5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2}) - 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3}) - 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2}))

צעדי פתרון

\frac{d}{d θ} ((A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ)))

(A) (5 π r (θ) (θ)^{2} - 4 r (θ) (θ)^{2} θ - 4 r (θ) (θ)^{2} sin (θ))

פשט את:

A (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ))

= \frac{d}{d θ} (A (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ)))

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

:הוצא את הקבוע

= A \frac{d}{d θ} (5 π θ^{2} r (θ) - 4 θ^{3} r (θ) - 4 θ^{2} sin (θ) r (θ))

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

:השתמש בחוק החיבור

= A (\frac{d}{d θ} (5 π θ^{2} r (θ)) - \frac{d}{d θ} (4 θ^{3} r (θ)) - \frac{d}{d θ} (4 θ^{2} sin (θ) r (θ)))

\frac{d}{d θ} (5 π θ^{2} r (θ)) = 5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2})

\frac{d}{d θ} (4 θ^{3} r (θ)) = 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3})

\frac{d}{d θ} (4 θ^{2} sin (θ) r (θ)) = 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2})

= A (5 π (2 θ r (θ) + \frac{d}{d θ} (r (θ)) θ^{2}) - 4 (3 θ^{2} r (θ) + \frac{d}{d θ} (r (θ)) θ^{3}) - 4 (2 θ sin (θ) r (θ) + (cos (θ) r (θ) + \frac{d}{d θ} (r (θ)) sin (θ)) θ^{2}))

e x p an d (8.3 - 5.8) \frac{d}{d x} (6.1 (v (x), 8.3))

e x p an d \frac{( 2 x - 1 ) ^{2}}{2 x}

e x p an d \frac{( s - s e ^{- 4 s} - 10 )}{( s ^{2} \cdot ( s ^{2} + 2 s ) )}

e x p an d - \frac{5}{( 2 x - 1 ) ^{2}}

s im pl i f y (5 x - 1 + x + 3)^{2}