What is the tangent of f(x)=sin(pix^2)-pix,\at x=1 ?

The tangent of f(x)=sin(pix^2)-pix,\at x=1 is y=-3pix+2pi

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tangent of f(x)=sin(pix^2)-pix,\at x=1

tangent of

f (x) = sin (π x^{2}) - π x, at x = 1

y = - 3 π x + 2 π

Solution steps

Find the tangent point:

(1, - π)

Find the slope of

f (x) = sin (π x^{2}) - π x : \frac{df ( x )}{d x} = cos (π x^{2}) \cdot 2 π x - π

Compute the slope of

f (x) = sin (π x^{2}) - π x

x = 1 : m = - 3 π

Find the line with slope m=

- 3 π

and passing through

(1, - π) : y = - 3 π x + 2 π

y = - 3 π x + 2 π

y = x^{3} - 8 x^{2} + 21 x + 2

y = \frac{90}{x ^{5}}

2 y^{^{'}} - y = 4 a sin (3 t)

G (s) = \frac{8 s ^{2}}{7 e ^{s} + s}

\frac{( x ^{3} - 6 x ) ^{9}}{9}

What is the tangent of f(x)=sin(pix^2)-pix,\at x=1 ?
The tangent of f(x)=sin(pix^2)-pix,\at x=1 is y=-3pix+2pi