What is the tangent of f(x)=5e^x+7x,\at x=0 ?

The tangent of f(x)=5e^x+7x,\at x=0 is y=12x+5

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tangent of f(x)=5e^x+7x,\at x=0

tangent of

f (x) = 5 e^{x} + 7 x, at x = 0

y = 12 x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = 5 e^{x} + 7 x : \frac{df ( x )}{d x} = 5 e^{x} + 7

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 = 12

Find the line with slope m=

12

and passing through

(0, 5) : y = 12 x + 5

y = 12 x + 5

\frac{\partial}{\partial t} (e^{2 x} - x cos (t))

x \to 0.9 lim (x^{2} - 1)

t an g e n t y = 13 x e^{x}

\frac{d y}{d x} - \frac{2 x ^{3} y}{x ^{4} + 3} = x^{3}

\frac{d}{d x} (2 A x + B)

What is the tangent of f(x)=5e^x+7x,\at x=0 ?
The tangent of f(x)=5e^x+7x,\at x=0 is y=12x+5