What is the tangent of f(x)=sin(x^2) ?

The tangent of f(x)=sin(x^2) is y=cos(a_{0}^2)*2a_{0}x+sin(a_{0}^2)-2a_{0}^2cos(a_{0}^2)

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tangent of f(x)=sin(x^2)

tangent of

f (x) = sin (x^{2})

y = cos (a_{0}^{2}) \cdot 2 a_{0} x + sin (a_{0}^{2}) - 2 a_{0}^{2} cos (a_{0}^{2})

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, sin (a_{0}^{2}))

Find the slope of

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

Compute the slope of

f (x) = sin (x^{2})

x = a_{0} : m = cos (a_{0}^{2}) \cdot 2 a_{0}

Find the line with slope m=

cos (a_{0}^{2}) 2 a_{0}

and passing through

(a_{0}, sin (a_{0}^{2})) : y = cos (a_{0}^{2}) \cdot 2 a_{0} x + sin (a_{0}^{2}) - 2 a_{0}^{2} cos (a_{0}^{2})

y = cos (a_{0}^{2}) \cdot 2 a_{0} x + sin (a_{0}^{2}) - 2 a_{0}^{2} cos (a_{0}^{2})

\int_{1}^{2} \frac{x ^{6} + 4 x ^{8}}{x ^{5}} d x

\int x^{2} + 6 x + 2 d x

\frac{\partial}{\partial x} (x + 1 + 1)

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

(2 x^{2} - 1)^{- 1}

What is the tangent of f(x)=sin(x^2) ?
The tangent of f(x)=sin(x^2) is y=cos(a_{0}^2)*2a_{0}x+sin(a_{0}^2)-2a_{0}^2cos(a_{0}^2)