What is the tangent of x^2-tan(x) ?

The tangent of x^2-tan(x) is y=(2a_{0}-sec^2(a_{0}))x-a_{0}^2+a_{0}sec^2(a_{0})-tan(a_{0})

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tangent of x^2-tan(x)

tangent of

x^{2} - tan (x)

y = (2 a_{0} - sec^{2} (a_{0})) x - a_{0}^{2} + a_{0} sec^{2} (a_{0}) - tan (a_{0})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - tan (a_{0}))

Find the slope of

y = x^{2} - tan (x) : \frac{d y}{d x} = 2 x - sec^{2} (x)

Compute the slope of

y = x^{2} - tan (x)

x = a_{0} : m = 2 a_{0} - sec^{2} (a_{0})

Find the line with slope m=

2 a_{0} - sec^{2} (a_{0})

and passing through

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

y = (2 a_{0} - sec^{2} (a_{0})) x - a_{0}^{2} + a_{0} sec^{2} (a_{0}) - tan (a_{0})

\frac{d y}{d x}, x^{2} - y^{2} = 64

\frac{d}{d x} (sec (\frac{π}{4}))

cos (\frac{x + 1}{x - 1})

F (x) = \frac{3}{1 + x ^{2}}

\int 3 x - 5 d x

What is the tangent of x^2-tan(x) ?
The tangent of x^2-tan(x) is y=(2a_{0}-sec^2(a_{0}))x-a_{0}^2+a_{0}sec^2(a_{0})-tan(a_{0})