What is the tangent of f(x)=1+5e^x,\at x=1 ?

The tangent of f(x)=1+5e^x,\at x=1 is y=5ex+1

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tangent of f(x)=1+5e^x,\at x=1

tangent of

f (x) = 1 + 5 e^{x}, at x = 1

y = 5 e x + 1

Solution steps

Find the tangent point:

(1, 1 + 5 e)

Find the slope of

f (x) = 1 + 5 e^{x} : \frac{df ( x )}{d x} = 5 e^{x}

Compute the slope of

f (x) = 1 + 5 e^{x}

x = 1 : m = 5 e

Find the line with slope m=

5 e

and passing through

(1, 1 + 5 e) : y = 5 e x + 1

y = 5 e x + 1

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

x y + y^{2} + x^{2} - x^{2} y^{^{'}} = 0

\int sin (105 x) sin (45 x) d x

y = 2 x^{3} + 3 x^{2} - 12 x + 8

\frac{d y}{d x} - 2 y = 3 x

What is the tangent of f(x)=1+5e^x,\at x=1 ?
The tangent of f(x)=1+5e^x,\at x=1 is y=5ex+1