What is the tangent of y=tan(12x^2),\at x=sqrt(pi) ?

The tangent of y=tan(12x^2),\at x=sqrt(pi) is y=24sqrt(pi)x-24pi

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tangent of y=tan(12x^2),\at x=sqrt(pi)

tangent of

y = tan (12 x^{2}), at x = π

y = 24 π x - 24 π

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

y = tan (12 x^{2}) : \frac{d y}{d x} = sec^{2} (12 x^{2}) \cdot 24 x

Compute the slope of

y = tan (12 x^{2})

x = π : m = 24 π

Find the line with slope m=

24 π

and passing through

(π, 0) : y = 24 π x - 24 π

y = 24 π x - 24 π

(x^{2})^{2}

sec^{2} (0)

n = 1 \sum \infty \frac{2}{n ^{6}}

\int \frac{x ^{4} - 3 x ^{2} + 4 x}{5 x} d x

x \to \infty lim (\frac{e ^{x}}{1 - e ^{- x}})

What is the tangent of y=tan(12x^2),\at x=sqrt(pi) ?
The tangent of y=tan(12x^2),\at x=sqrt(pi) is y=24sqrt(pi)x-24pi