What is the tangent of y=e^{3x},\at x= 1/3 ln(3) ?

The tangent of y=e^{3x},\at x= 1/3 ln(3) is y=9x+3-3ln(3)

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tangent of y=e^{3x},\at x= 1/3 ln(3)

tangent of

y = e^{3 x}, at x = \frac{1}{3} ln (3)

y = 9 x + 3 - 3 ln (3)

Solution steps

Find the tangent point:

(\frac{1}{3} ln (3), 3)

Find the slope of

y = e^{3 x} : \frac{d y}{d x} = e^{3 x} \cdot 3

Compute the slope of

y = e^{3 x}

x = \frac{1}{3} ln (3) : m = 9

Find the line with slope m=

9

and passing through

(\frac{1}{3} ln (3), 3) : y = 9 x + 3 - 3 ln (3)

y = 9 x + 3 - 3 ln (3)

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

6 - x^{2}

\int_{- y}^{y} x d x

\frac{d}{d x} (2 x e^{- 3 x})

\frac{d}{d x} (x^{6} (x - 4)^{5})

What is the tangent of y=e^{3x},\at x= 1/3 ln(3) ?
The tangent of y=e^{3x},\at x= 1/3 ln(3) is y=9x+3-3ln(3)