What is the tangent of e^t+3,\at t=0 ?

The tangent of e^t+3,\at t=0 is y=t+4

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tangent of e^t+3,\at t=0

tangent of

e^{t} + 3, at t = 0

y = t + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

y = e^{t} + 3 : \frac{d y}{d t} = e^{t}

Compute the slope of

y = e^{t} + 3

t = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 4) : y = t + 4

y = t + 4

y^{^{''}} + 4 y = sin (t)

x^{- 2}, 6

y^{2} = x^{3}, y^{2} = x

\int \frac{- 9 x ^{2} + 2}{x ^{3} + 9 x} d x

\int \frac{1}{x ^{2} x ^{2} + 49} d x