What is the tangent of y=9e^x+x ?

The tangent of y=9e^x+x is y=(9e^{a_{0}}+1)x-9a_{0}e^{a_{0}}+9e^{a_{0}}

Popular >

tangent of y=9e^x+x

tangent of

y = 9 e^{x} + x

y = (9 e^{a_{0}} + 1) x - 9 a_{0} e^{a_{0}} + 9 e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 e^{a_{0}} + a_{0})

Find the slope of

y = 9 e^{x} + x : \frac{d y}{d x} = 9 e^{x} + 1

Compute the slope of

y = 9 e^{x} + x

x = a_{0} : m = 9 e^{a_{0}} + 1

Find the line with slope m=

9 e^{a_{0}} + 1

and passing through

(a_{0}, 9 e^{a_{0}} + a_{0}) : y = (9 e^{a_{0}} + 1) x - 9 a_{0} e^{a_{0}} + 9 e^{a_{0}}

y = (9 e^{a_{0}} + 1) x - 9 a_{0} e^{a_{0}} + 9 e^{a_{0}}

\frac{d y}{d x} = e^{2 x - 3 y}

\frac{\partial}{\partial x} (7 x cos (x) cos (y))

\frac{d y}{d t} = 8 - \frac{y}{200 + 3 t}

7 e^{x} cos (x)

f (x) = - 2 x e^{x^{2}}, at x = 2

What is the tangent of y=9e^x+x ?
The tangent of y=9e^x+x is y=(9e^{a_{0}}+1)x-9a_{0}e^{a_{0}}+9e^{a_{0}}