What is the d/(dq)(-0.1q^2+1.2q+46.4) ?

The d/(dq)(-0.1q^2+1.2q+46.4) is -0.2q+1.2

What is the first d/(dq)(-0.1q^2+1.2q+46.4) ?

The first d/(dq)(-0.1q^2+1.2q+46.4) is -0.2q+1.2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d q} (- 0.1 q^{2} + 1.2 q + 46.4)

- 0.2 q + 1.2

Solution steps

\frac{d}{d q} (- 0.1 q^{2} + 1.2 q + 46.4)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d q} (0.1 q^{2}) + \frac{d}{d q} (1.2 q) + \frac{d}{d q} (46.4)

\frac{d}{d q} (0.1 q^{2}) = 0.2 q

\frac{d}{d q} (1.2 q) = 1.2

\frac{d}{d q} (46.4) = 0

= - 0.2 q + 1.2 + 0

Simplify

= - 0.2 q + 1.2

n = 1 \sum \infty (\frac{9}{8})^{n}

\frac{d y}{d x}, x y + 3 = y

\int \frac{( x ^{7} )}{1 + x ^{16}} d x

\frac{\partial}{\partial x} (2 x^{3} y - 3 y^{2} z)

2 4 - x

What is the d/(dq)(-0.1q^2+1.2q+46.4) ?
The d/(dq)(-0.1q^2+1.2q+46.4) is -0.2q+1.2
What is the first d/(dq)(-0.1q^2+1.2q+46.4) ?
The first d/(dq)(-0.1q^2+1.2q+46.4) is -0.2q+1.2