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

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Answer 1 · 2018-04-02T16:59:37

Hence equation of tangent is y=1

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

Slope of the tangent is

$\frac{d y}{d x} = 3 {cos}^{2} x^{2} (- {sin x}^{2}) (2 x)$ $dy/dx=3cos^2x^2(-sinx^2)(2x)$

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

at x=0

$m = - 6 \times 0 \times {cos}^{2} 0^{2} {sin 0}^{2}$ $m=-6xx0xxcos^2 0^2sin0^2$

$m = 0$ $m=0$

point of tangency is

$y = {cos}^{3} x^{2} = {cos}^{3} 0^{2} = {cos}^{3} 0 = {(cos 0)}^{2} = 1^{2} = 1$ $y=cos^3x^2=cos^3 0^2=cos^3 0=(cos0)^2=1^2=1$

Hence equation of tangent is y=1

What is the equation of the tangent line of f(x)=cos^3x^2 f(x)=cos3x2f(x)=cos^3x^2 at x=0x=0x=0?