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

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

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

tangent of

f (x) = 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

f (x) = e^{3 x} : \frac{df ( x )}{d x} = e^{3 x} \cdot 3

Compute the slope of

f (x) = 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{d}{d x} (sin (x) ln (2 x))

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

\int \frac{cos ( θ )}{1 + sin ^{2} ( θ )} d θ

\int (1 + y e^{x y}) d x

\int \frac{1}{x ( ln ( x ) ) ^{2}} d x

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