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

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

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

tangent of

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

y = 15 x + 3 - 3 ln (3)

Solution steps

Find the tangent point:

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

Find the slope of

y = e^{5 x} : \frac{d y}{d x} = e^{5 x} \cdot 5

Compute the slope of

y = e^{5 x}

x = \frac{1}{5} ln (3) : m = 15

Find the line with slope m=

15

and passing through

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

y = 15 x + 3 - 3 ln (3)

\int (x + 1) ln (x + 1) d x

f (x) = (1 + 9 x^{2}) (x - x^{2})

\int (e^{3 x} + 2)^{2} d x

\int \frac{1}{x + 5 x + 6} d x

\frac{d}{d x} (sin (\frac{π}{6}))

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