What is the tangent of f(x)=e^x,\at x=ln(3) ?

The tangent of f(x)=e^x,\at x=ln(3) is y=3x+3-3ln(3)

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tangent of f(x)=e^x,\at x=ln(3)

tangent of

f (x) = e^{x}, at x = ln (3)

y = 3 x + 3 - 3 ln (3)

Solution steps

Find the tangent point:

(ln (3), 3)

Find the slope of

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

Compute the slope of

f (x) = e^{x}

x = ln (3) : m = 3

Find the line with slope m=

3

and passing through

(ln (3), 3) : y = 3 x + 3 - 3 ln (3)

y = 3 x + 3 - 3 ln (3)

\int \frac{x ^{3}}{x ^{2} + 4} d x

5 sin (3 t - \frac{π}{4})

e^{x}, e^{- x}, [0, 1]

\frac{\partial}{\partial x} ((x^{3} - y^{2})^{- 1})

x \to 0 + lim ((1 + x)^{(\frac{1}{x})})

What is the tangent of f(x)=e^x,\at x=ln(3) ?
The tangent of f(x)=e^x,\at x=ln(3) is y=3x+3-3ln(3)