What is the tangent of y=x(x+1)^5,(1,32) ?

The tangent of y=x(x+1)^5,(1,32) is y=112x-80

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x(x+1)^5,(1,32)

tangent of

y = x (x + 1)^{5}, (1, 32)

y = 112 x - 80

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 32) : m = 112

Find the line with slope m=

112

and passing through

(1, 32) : y = 112 x - 80

y = 112 x - 80

\int_{\frac{3}{4}}^{1} 2 x d x

\frac{d y}{d x} + y^{9} x + 6 y = 0

y^{^{''}} + 2 y^{^{'}} + y = e - t ln (t)

\frac{\partial}{\partial y} (\frac{( x - y )}{x + y})

\int_{- 1}^{2} (x^{2} - 1) d x

What is the tangent of y=x(x+1)^5,(1,32) ?
The tangent of y=x(x+1)^5,(1,32) is y=112x-80