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

The tangent of y=e^{2x},\at x= 1/2 ln(2) is y=4x+2-2ln(2)

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

tangent of

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

y = 4 x + 2 - 2 ln (2)

Solution steps

Find the tangent point:

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

Find the slope of

y = e^{2 x} : \frac{d y}{d x} = e^{2 x} \cdot 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

(\frac{1}{2} ln (2), 2) : y = 4 x + 2 - 2 ln (2)

y = 4 x + 2 - 2 ln (2)

x \to - 1 lim (\frac{∣ x ∣}{x + 1})

\int x arctan (8 x) d x

f (x) = cos^{2} (4 x)

\frac{d}{d x} ((6 x + e^{- x^{2}})^{5})

x \to 0 lim (\frac{2 x}{sin ( x )})

What is the tangent of y=e^{2x},\at x= 1/2 ln(2) ?
The tangent of y=e^{2x},\at x= 1/2 ln(2) is y=4x+2-2ln(2)