What is the tangent of 8e^x+x ?

The tangent of 8e^x+x is y=(8e^{a_{0}}+1)x-8a_{0}e^{a_{0}}+8e^{a_{0}}

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tangent of 8e^x+x

tangent of

8 e^{x} + x

y = (8 e^{a_{0}} + 1) x - 8 a_{0} e^{a_{0}} + 8 e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 e^{a_{0}} + a_{0})

Find the slope of

y = 8 e^{x} + x : \frac{d y}{d x} = 8 e^{x} + 1

Compute the slope of

y = 8 e^{x} + x

x = a_{0} : m = 8 e^{a_{0}} + 1

Find the line with slope m=

8 e^{a_{0}} + 1

and passing through

(a_{0}, 8 e^{a_{0}} + a_{0}) : y = (8 e^{a_{0}} + 1) x - 8 a_{0} e^{a_{0}} + 8 e^{a_{0}}

y = (8 e^{a_{0}} + 1) x - 8 a_{0} e^{a_{0}} + 8 e^{a_{0}}

x \to - 1 lim (\frac{x}{1 - x ^{4}})

\int_{0}^{1} (1 + r)^{3} d r

s \cdot + 5

x \to 2 lim (\frac{x - 1}{x ^{2} - 1})

\int x^{4} (ln (x))^{2} d x

What is the tangent of 8e^x+x ?
The tangent of 8e^x+x is y=(8e^{a_{0}}+1)x-8a_{0}e^{a_{0}}+8e^{a_{0}}