# What is the equation of the tangent line of f(x)=cos^3x^2  at x=0?

Hence equation of tangent is y=1

#### Explanation:

Let
$y = {\cos}^{3} {x}^{2}$

Slope of the tangent is

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\cos}^{2} {x}^{2} \left(- \sin {x}^{2}\right) \left(2 x\right)$

$m = - 6 x {\cos}^{2} {x}^{2} \sin {x}^{2}$

at x=0

$m = - 6 \times 0 \times {\cos}^{2} {0}^{2} \sin {0}^{2}$

$m = 0$

point of tangency is

$y = {\cos}^{3} {x}^{2} = {\cos}^{3} {0}^{2} = {\cos}^{3} 0 = {\left(\cos 0\right)}^{2} = {1}^{2} = 1$

Hence equation of tangent is y=1