What is the tangent of 6x^2-5 ?

The tangent of 6x^2-5 is y=12a_{0}x-6a_{0}^2-5

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tangent of 6x^2-5

tangent of

6 x^{2} - 5

y = 12 a_{0} x - 6 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} - 5)

Find the slope of

y = 6 x^{2} - 5 : \frac{d y}{d x} = 12 x

Compute the slope of

y = 6 x^{2} - 5

x = a_{0} : m = 12 a_{0}

Find the line with slope m=

12 a_{0}

and passing through

(a_{0}, 6 a_{0}^{2} - 5) : y = 12 a_{0} x - 6 a_{0}^{2} - 5

y = 12 a_{0} x - 6 a_{0}^{2} - 5

\frac{\partial}{\partial x} (sin (5 x + 5 y))

n = 1 \sum \infty sin^{2} (n)

f (x) = x, (81, 9)

3 e^{- 5 t} + 6 e^{- 3 t}

\frac{d}{d t} (\frac{4}{t ^{2}} - \frac{5}{t} - \frac{9}{5})