What is the tangent of 3^x ?

The tangent of 3^x is y=3^{a_{0}}ln(3)x+3^{a_{0}}-3^{a_{0}}ln(3)a_{0}

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tangent of 3^x

tangent of

3^{x}

y = 3^{a_{0}} ln (3) x + 3^{a_{0}} - 3^{a_{0}} ln (3) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3^{a_{0}})

Find the slope of

y = 3^{x} : \frac{d y}{d x} = 3^{x} ln (3)

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 = 3^{a_{0}} ln (3)

Find the line with slope m=

3^{a_{0}} ln (3)

and passing through

(a_{0}, 3^{a_{0}}) : y = 3^{a_{0}} ln (3) x + 3^{a_{0}} - 3^{a_{0}} ln (3) a_{0}

y = 3^{a_{0}} ln (3) x + 3^{a_{0}} - 3^{a_{0}} ln (3) a_{0}

y^{^{'}} = - \frac{y}{x} + \frac{e ^{x}}{x}

\frac{d y}{d x} = \frac{( x - y )}{x + 2 y}

\int_{0}^{1} 2 (1 + x)^{7} d x

y = - 4 sin (x)

\frac{d}{d t} (2^{t^{3}})

What is the tangent of 3^x ?
The tangent of 3^x is y=3^{a_{0}}ln(3)x+3^{a_{0}}-3^{a_{0}}ln(3)a_{0}