What is the tangent of x^2+5 ?

The tangent of x^2+5 is y=2a_{0}x-a_{0}^2+5

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tangent of x^2+5

tangent of

x^{2} + 5

y = 2 a_{0} x - a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 5)

Find the slope of

y = x^{2} + 5 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 5

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 5) : y = 2 a_{0} x - a_{0}^{2} + 5

y = 2 a_{0} x - a_{0}^{2} + 5

\frac{d}{d x} (\frac{x ^{3}}{3} + 7 x^{2} + 50 x)

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\int \frac{e ^{x - y}}{1 + e ^{x}}

\frac{d}{d x} (lo g_{3} (3 x - 7))