What is the tangent of 5x^2 ?

The tangent of 5x^2 is y=10a_{0}x-5a_{0}^2

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tangent of 5x^2

tangent of

5 x^{2}

y = 10 a_{0} x - 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 5 x^{2} : \frac{d y}{d x} = 10 x

Compute the slope of

y = 5 x^{2}

x = a_{0} : m = 10 a_{0}

Find the line with slope m=

10 a_{0}

and passing through

(a_{0}, 5 a_{0}^{2}) : y = 10 a_{0} x - 5 a_{0}^{2}

y = 10 a_{0} x - 5 a_{0}^{2}

n = 1 \sum \infty (- 3)^{3 n - 1}

\frac{d}{d x} (ln (\frac{x ^{3} + 1}{x ^{2} + 1}))

\int 3 e^{3 t} d t

\frac{x}{x + 1}

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x