What is the tangent of y=e^{3x},\at x= 1/3 ln(5) ?

The tangent of y=e^{3x},\at x= 1/3 ln(5) is y=15x+5-5ln(5)

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tangent of y=e^{3x},\at x= 1/3 ln(5)

tangent of

y = e^{3 x}, at x = \frac{1}{3} ln (5)

y = 15 x + 5 - 5 ln (5)

Solution steps

Find the tangent point:

(\frac{1}{3} ln (5), 5)

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 (5) : m = 15

Find the line with slope m=

15

and passing through

(\frac{1}{3} ln (5), 5) : y = 15 x + 5 - 5 ln (5)

y = 15 x + 5 - 5 ln (5)

\frac{\partial}{\partial x} (\frac{5 x}{2 y})

\int_{0}^{1} (e^{- 0.2 t}) d t

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

\int arcsin (9 x) d x

y^{^{'}} + 8 y - 6 x = 0

What is the tangent of y=e^{3x},\at x= 1/3 ln(5) ?
The tangent of y=e^{3x},\at x= 1/3 ln(5) is y=15x+5-5ln(5)