What is the tangent of f(x)=e^{2x},(0,1) ?

The tangent of f(x)=e^{2x},(0,1) is y=2x+1

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tangent of f(x)=e^{2x},(0,1)

tangent of

f (x) = e^{2 x}, (0, 1)

y = 2 x + 1

Solution steps

Find the slope of

f (x) = e^{2 x} : \frac{df ( x )}{d x} = e^{2 x} \cdot 2

Compute the slope of

f (x) = e^{2 x}

(0, 1) : m = 2

Find the line with slope m=

2

and passing through

(0, 1) : y = 2 x + 1

y = 2 x + 1

n = 1 \sum \infty (- 1)^{n} n^{- 1}

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'}} = 8 x + y, y (0) = 6

\frac{\partial}{\partial z} (e^{x} + yz)

f (x)

What is the tangent of f(x)=e^{2x},(0,1) ?
The tangent of f(x)=e^{2x},(0,1) is y=2x+1