Question

The slope of tangent of the curve y = 2x2 + 3 sin x at x = 0 is:​

Answer 1

EXPLANATION.

Slope of tangent of the curve,

⇒ y = 2x² + 3 sin(x). at x = 0

As we know that,

Slope of the tangent = dy/dx.

Differentiate w.r.t x, we get.

⇒ dy/dx = d(2x² + 3 sin(x)/dx.

⇒ dy/dx = 4x + 3 cos(x).

Slope = M₁M₂ = -1.

⇒ (4x + 3 cos(x)).M₂ = -1.

⇒ M₂ = -1/(4x + 3 cos(x)).

Put x = 0 in equation, we get.

⇒ M₂ = -1/(4(0) + 3 cos(0)).

⇒ M₂ = -1/3.

                                                                                                                                 

MORE INFORMATION.

Equation of tangent and normal in ''parametric form''.

If x = f(t) and y = g(t) then,

(1) =  Equation of tangent is,

⇒ (y - g(t)) = g'(t)/f'(t) (x - f(t)).

(2) = Equation of normal,

⇒ (y - g(t)) = - f'(x)/g'(x) (x - f(t)).

Answer 2

Step-by-step explanation:

$slop \: of \: trangel \: of \: the \: curve \\ \\ = \: y \: = 2x {}^{2} + 3 \sin(x) .at \: x = \\ answer \: know \: that \: \\ slop \: of \: the \\ different \: \\ trangle \: = d \: y \: \frac{dy}{dx} differentinde \: w.r.dxwe \: get \: \\ \frac{dy}{dx} = d( {2x}^{2} + 3 \sin(x) \frac{?}{dx \\ \\ slop \: = m1m2 = - (4x + 3 \cos(x) .m2 = - 1. \\ \\ m2 = - \frac{1}{(4x + 3 \cos(x) } ) \\ put \: x \: = 0 \: in \: equation \: we \: get \: m \: 2 = \frac{1}{4 \cos(4) } + 3 \cos(0) . \: m2 = \frac{1}{3} .$