What is the tangent of y=x^3+4x,(1,5) ?

The tangent of y=x^3+4x,(1,5) is y=7x-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^3+4x,(1,5)

tangent of

y = x^{3} + 4 x, (1, 5)

y = 7 x - 2

Solution steps

Find the slope of

y = x^{3} + 4 x : \frac{d y}{d x} = 3 x^{2} + 4

Compute the slope of

y = x^{3} + 4 x

(1, 5) : m = 7

Find the line with slope m=

7

and passing through

(1, 5) : y = 7 x - 2

y = 7 x - 2

x \to - 10 lim (∣ x + 10 ∣ + 3)

f (x) = \frac{1}{x ^{5}} - x^{2} - 2 x, at x = 1

f (x) = \frac{x ^{2}}{e ^{x}}

x \to 0 lim (\frac{2}{x} - \frac{2}{x ^{2} + x})

\frac{d}{d x} ((3 x - 4)^{5})