What is the tangent of y=2^x ?

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

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tangent of y=2^x

tangent of

y = 2^{x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2^{x} : \frac{d y}{d x} = 2^{x} ln (2)

Compute the slope of

y = 2^{x}

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

Find the line with slope m=

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

and passing through

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

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

\frac{\partial}{\partial z} (3 x z^{3} e^{y^{2}})

\frac{\partial}{\partial z} (x z + x yz)

\int (x^{11} e^{x^{12}}) d x

\int \frac{3 x ^{2} + 2}{x ^{2}} d x

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

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