What is the d/(dθ)(5sin(2θ)) ?

The d/(dθ)(5sin(2θ)) is 10cos(2θ)

What is the first d/(dθ)(5sin(2θ)) ?

The first d/(dθ)(5sin(2θ)) is 10cos(2θ)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d θ} (5 sin (2 θ))

10 cos (2 θ)

Solution steps

\frac{d}{d θ} (5 sin (2 θ))

Take the constant out:

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

= 5 \frac{d}{d θ} (sin (2 θ))

Apply the chain rule:

cos (2 θ) \frac{d}{d θ} (2 θ)

= cos (2 θ) \frac{d}{d θ} (2 θ)

\frac{d}{d θ} (2 θ) = 2

= 5 cos (2 θ) \cdot 2

Simplify

= 10 cos (2 θ)

y^{^{''}} + 3 y = 0

\int \frac{x ^{2} + 2 x + 3}{x ^{3} - x} d x

\int_{2}^{3} \frac{1}{x - 1} d x

x \to - 4 lim (6 x + 11)

\frac{d}{d x} (x^{2} \cdot tan (\frac{1}{x}))