What is the derivative of (Ax^2+Bx+C(e^{4x})) ?

The derivative of (Ax^2+Bx+C(e^{4x})) is (2Ax+B)e^{4x}+e^{4x}*4(Ax^2+Bx+C)

What is the first derivative of (Ax^2+Bx+C(e^{4x})) ?

The first derivative of (Ax^2+Bx+C(e^{4x})) is (2Ax+B)e^{4x}+e^{4x}*4(Ax^2+Bx+C)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} ((A x^{2} + B x + C) (e^{4 x}))

(2 A x + B) e^{4 x} + e^{4 x} \cdot 4 (A x^{2} + B x + C)

Solution steps

\frac{d}{d x} ((A x^{2} + B x + C) (e^{4 x}))

Treat

A, B, C

as constants

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = A x^{2} + B x + C, g = e^{4 x}

= \frac{d}{d x} (A x^{2} + B x + C) e^{4 x} + \frac{d}{d x} (e^{4 x}) (A x^{2} + B x + C)

\frac{d}{d x} (A x^{2} + B x + C) = 2 A x + B

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

= (2 A x + B) e^{4 x} + e^{4 x} \cdot 4 (A x^{2} + B x + C)

y^{^{'}} = \frac{2 y ^{4} - 2}{4 x y ^{3}}

n = 3 \sum \infty 1 + \frac{1}{n ^{2}}

x \cdot e^{x}

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

(∣ x ∣)^{^{'}}

What is the derivative of (Ax^2+Bx+C(e^{4x})) ?
The derivative of (Ax^2+Bx+C(e^{4x})) is (2Ax+B)e^{4x}+e^{4x}*4(Ax^2+Bx+C)
What is the first derivative of (Ax^2+Bx+C(e^{4x})) ?
The first derivative of (Ax^2+Bx+C(e^{4x})) is (2Ax+B)e^{4x}+e^{4x}*4(Ax^2+Bx+C)