What is the tangent of y=tan(18x^2),\at x=sqrt(pi) ?

The tangent of y=tan(18x^2),\at x=sqrt(pi) is y=36sqrt(pi)x-36pi

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tangent of y=tan(18x^2),\at x=sqrt(pi)

tangent of

y = tan (18 x^{2}), at x = π

y = 36 π x - 36 π

Solution steps

Find the tangent point:

(π, 0)

Find the slope of

y = tan (18 x^{2}) : \frac{d y}{d x} = sec^{2} (18 x^{2}) \cdot 36 x

Compute the slope of

y = tan (18 x^{2})

x = π : m = 36 π

Find the line with slope m=

36 π

and passing through

(π, 0) : y = 36 π x - 36 π

y = 36 π x - 36 π

\int 2 cos (7 x - 4) d x

y^{^{''}} + 9 y = csc^{2} (3 t)

y^{^{'}} = 2 y - t^{2}

\int_{0}^{2} sin (x^{2}) d x

\frac{\partial}{\partial x} (- x y z^{2} + 16)

What is the tangent of y=tan(18x^2),\at x=sqrt(pi) ?
The tangent of y=tan(18x^2),\at x=sqrt(pi) is y=36sqrt(pi)x-36pi