What is the tangent of f(x)=e^{5x},\at x= 1/5 ln(2) ?

The tangent of f(x)=e^{5x},\at x= 1/5 ln(2) is y=10x+2-2ln(2)

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tangent of f(x)=e^{5x},\at x= 1/5 ln(2)

tangent of

f (x) = e^{5 x}, at x = \frac{1}{5} ln (2)

y = 10 x + 2 - 2 ln (2)

Solution steps

Find the tangent point:

(\frac{1}{5} ln (2), 2)

Find the slope of

f (x) = e^{5 x} : \frac{df ( x )}{d x} = e^{5 x} \cdot 5

Compute the slope of

f (x) = e^{5 x}

x = \frac{1}{5} ln (2) : m = 10

Find the line with slope m=

10

and passing through

(\frac{1}{5} ln (2), 2) : y = 10 x + 2 - 2 ln (2)

y = 10 x + 2 - 2 ln (2)

\frac{\partial}{\partial x} (96 - 5 x^{2} - 3 y^{2})

f (x) = x^{3} + 1

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

\frac{\partial}{\partial y} (∣ y ∣ + \frac{1}{2 a} ∣ x - y ∣^{2})

\int \frac{1}{4 x + 1} d x

What is the tangent of f(x)=e^{5x},\at x= 1/5 ln(2) ?
The tangent of f(x)=e^{5x},\at x= 1/5 ln(2) is y=10x+2-2ln(2)