What is the tangent of f(x)=x2^x ?

The tangent of f(x)=x2^x is y=(2^{a_{0}}+ln(2)*2^{a_{0}}a_{0})x-ln(2)*2^{a_{0}}a_{0}^2

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tangent of f(x)=x2^x

tangent of

f (x) = x 2^{x}

y = (2^{a_{0}} + ln (2) \cdot 2^{a_{0}} a_{0}) x - ln (2) \cdot 2^{a_{0}} a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} \cdot 2^{a_{0}})

Find the slope of

f (x) = x 2^{x} : \frac{df ( x )}{d x} = 2^{x} + ln (2) \cdot 2^{x} x

Compute the slope of

f (x) = x 2^{x}

x = a_{0} : m = 2^{a_{0}} + ln (2) \cdot 2^{a_{0}} a_{0}

Find the line with slope m=

2^{a_{0}} + ln (2) 2^{a_{0}} a_{0}

and passing through

(a_{0}, a_{0} 2^{a_{0}}) : y = (2^{a_{0}} + ln (2) \cdot 2^{a_{0}} a_{0}) x - ln (2) \cdot 2^{a_{0}} a_{0}^{2}

y = (2^{a_{0}} + ln (2) \cdot 2^{a_{0}} a_{0}) x - ln (2) \cdot 2^{a_{0}} a_{0}^{2}

(- 2.3) (2.2)

x \to 0 - lim (x \cdot arctan (\frac{1}{x}))

\frac{d y}{d x} = \frac{2 x}{1 + 2 y}, y (2) = 0

n = 1 \sum \infty (- 1)^{n - 1} \frac{2 ^{n}}{n ^{4}}

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

What is the tangent of f(x)=x2^x ?
The tangent of f(x)=x2^x is y=(2^{a_{0}}+ln(2)*2^{a_{0}}a_{0})x-ln(2)*2^{a_{0}}a_{0}^2