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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

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