What is the tangent of f(x)=x+4e^x ?

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

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tangent of f(x)=x+4e^x

tangent of

f (x) = x + 4 e^{x}

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

Find the slope of

f (x) = x + 4 e^{x} : \frac{df ( x )}{d x} = 1 + 4 e^{x}

Compute the slope of

f (x) = x + 4 e^{x}

x = a_{0} : m = 1 + 4 e^{a_{0}}

Find the line with slope m=

1 + 4 e^{a_{0}}

and passing through

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

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

\frac{d}{d t} (a cos (wt))

y = 2 x^{2} + 8 x + 10

\frac{d}{d x} ((4 x^{2} + 5) (5 x - 3))

x \to 90 lim (cos (x))

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