What is the tangent of 4e^x+x ?

The tangent of 4e^x+x is y=(4e^{a_{0}}+1)x-4a_{0}e^{a_{0}}+4e^{a_{0}}

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tangent of 4e^x+x

tangent of

4 e^{x} + x

y = (4 e^{a_{0}} + 1) x - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 e^{a_{0}} + a_{0})

Find the slope of

y = 4 e^{x} + x : \frac{d y}{d x} = 4 e^{x} + 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4 e^{a_{0}} + 1

Find the line with slope m=

4 e^{a_{0}} + 1

and passing through

(a_{0}, 4 e^{a_{0}} + a_{0}) : y = (4 e^{a_{0}} + 1) x - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

y = (4 e^{a_{0}} + 1) x - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

\int_{0}^{9} \frac{1}{81 + t ^{2}} d t

x^{8}

\frac{d}{d x} (x (x - 2)^{2})

\frac{1}{s ^{2} + 4 s + 20}

y^{^{'}} = 1 0^{- 21} \cdot y^{3}

What is the tangent of 4e^x+x ?
The tangent of 4e^x+x is y=(4e^{a_{0}}+1)x-4a_{0}e^{a_{0}}+4e^{a_{0}}