What is the tangent of y=(1+5x)^8,(0,1) ?

The tangent of y=(1+5x)^8,(0,1) is y=40x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=(1+5x)^8,(0,1)

tangent of

y = (1 + 5 x)^{8}, (0, 1)

y = 40 x + 1

Solution steps

Find the slope of

y = (1 + 5 x)^{8} : \frac{d y}{d x} = 40 (1 + 5 x)^{7}

Compute the slope of

y = (1 + 5 x)^{8}

(0, 1) : m = 40

Find the line with slope m=

40

and passing through

(0, 1) : y = 40 x + 1

y = 40 x + 1

\frac{d}{d x} (n ln (0.4 n x + 0.6))

\frac{( x ^{2} )}{8 + x}

y = x^{2}, y = 2 - x^{2}

e^{x^{5}}

What is the tangent of y=(1+5x)^8,(0,1) ?
The tangent of y=(1+5x)^8,(0,1) is y=40x+1