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

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

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tangent of 9e^x+x

tangent of

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}}

x \to - 4 lim (\frac{x ^{2} - 2}{4 - x})

ln (5 x^{4})

\int \frac{x ^{2}}{4} d x

\frac{\partial}{\partial x} ((e^{2 y} - y cos (x y)) d x)

\int e^{x} 1 + e^{2 x} d x

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