What is the tangent of f(x)=e^{3x},(0,1) ?

The tangent of f(x)=e^{3x},(0,1) is y=3x+1

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tangent of f(x)=e^{3x},(0,1)

tangent of

f (x) = e^{3 x}, (0, 1)

y = 3 x + 1

Solution steps

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}

(0, 1) : m = 3

Find the line with slope m=

3

and passing through

(0, 1) : y = 3 x + 1

y = 3 x + 1

\frac{d}{d x} ((2 x + 5) (x + 1))

\frac{d}{d x} (\frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}})

y + y^{^{'}} = sin (t)

t^{9} cos (t)

h \to + 0 lim (\frac{∣ h ∣}{h})

What is the tangent of f(x)=e^{3x},(0,1) ?
The tangent of f(x)=e^{3x},(0,1) is y=3x+1