What is the tangent of y=3e^x+x,(0,3) ?

The tangent of y=3e^x+x,(0,3) is y=4x+3

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tangent of y=3e^x+x,(0,3)

tangent of

y = 3 e^{x} + x, (0, 3)

y = 4 x + 3

Solution steps

Find the slope of

y = 3 e^{x} + x : \frac{d y}{d x} = 3 e^{x} + 1

Compute the slope of

y = 3 e^{x} + x

(0, 3) : m = 4

Find the line with slope m=

4

and passing through

(0, 3) : y = 4 x + 3

y = 4 x + 3

\frac{d}{d x} (2 x^{2} - 8)

\frac{7}{( 7 - x ^{8} ) ^{\frac{3}{2}}}

(sin^{2} (x) + 1)^{5}

2 t - sin (2 t)

\int cos^{3} (x) d x