What is the tangent of 4^x ?

The tangent of 4^x is y=ln(2)*2^{2a_{0}+1}x+4^{a_{0}}-ln(2)*2^{2a_{0}+1}a_{0}

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tangent of 4^x

tangent of

4^{x}

y = ln (2) \cdot 2^{2 a_{0} + 1} x + 4^{a_{0}} - ln (2) \cdot 2^{2 a_{0} + 1} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 4^{x} : \frac{d y}{d x} = ln (2) \cdot 2^{2 x + 1}

Compute the slope of

y = 4^{x}

x = a_{0} : m = ln (2) \cdot 2^{2 a_{0} + 1}

Find the line with slope m=

ln (2) 2^{2 a_{0} + 1}

and passing through

(a_{0}, 4^{a_{0}}) : y = ln (2) \cdot 2^{2 a_{0} + 1} x + 4^{a_{0}} - ln (2) \cdot 2^{2 a_{0} + 1} a_{0}

y = ln (2) \cdot 2^{2 a_{0} + 1} x + 4^{a_{0}} - ln (2) \cdot 2^{2 a_{0} + 1} a_{0}

y = \frac{e ^{x} + e ^{- x}}{1 + e ^{2 x}}

\int - x^{\frac{5}{2}} d x

\frac{d y}{d x} - 2 x y^{2} = 0

\int ln ((6 x)^{2}) d x

\int \frac{1}{sin ^{2} ( x )} d x

What is the tangent of 4^x ?
The tangent of 4^x is y=ln(2)*2^{2a_{0}+1}x+4^{a_{0}}-ln(2)*2^{2a_{0}+1}a_{0}