What is the derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 ?

The derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 is (x^4)/7+x^3+1

What is the first derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 ?

The first derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 is (x^4)/7+x^3+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = \frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1

\frac{x ^{4}}{7} + x^{3} + 1

Solution steps

\frac{d}{d x} (\frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1)

Simplify

\frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1 : \frac{x ^{5}}{35} + \frac{x ^{4}}{4} + x + 1

= \frac{d}{d x} (\frac{x ^{5}}{35} + \frac{x ^{4}}{4} + x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{5}}{35}) + \frac{d}{d x} (\frac{x ^{4}}{4}) + \frac{d x}{d x} + \frac{d}{d x} (1)

\frac{d}{d x} (\frac{x ^{5}}{35}) = \frac{x ^{4}}{7}

\frac{d}{d x} (\frac{x ^{4}}{4}) = x^{3}

\frac{d x}{d x} = 1

\frac{d}{d x} (1) = 0

= \frac{x ^{4}}{7} + x^{3} + 1 + 0

Simplify

= \frac{x ^{4}}{7} + x^{3} + 1

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

\frac{d}{d x} (\frac{x ^{2} + 9}{x})

\frac{d}{d x} (x^{2} ln^{2} (x))

n = 0 \sum \infty e^{- \frac{5 n}{2}}

\frac{\partial}{\partial x} (x^{3} e^{y} + cos (y))

What is the derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 ?
The derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 is (x^4)/7+x^3+1
What is the first derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 ?
The first derivative of f(x)= 1/35 x^5+1/4 x^4+x+1 is (x^4)/7+x^3+1