Respuesta :
Answer:
slope of the tangent
[tex]\frac{d y}{d x} = \frac{cos(x+sin^{-1}(x) (1+\sqrt{1-x^{2}) } }{\sqrt{1-x^{2} } }[/tex]
The function y = f(x)
y(x) = 2 x + C
Step-by-step explanation:
Step(i):-
Given x= sin t ...(i)
Differentiating equation (i) with respective to 'x' , we get
[tex]\frac{d x}{d t} = cost[/tex]
Given y = sin ( t + sin (t)) ...(ii)
Differentiating equation (ii) with respective to 'x' , we get
[tex]\frac{d y}{d t} = cos (t + sin t ) (1 + cos t)[/tex]
Step(ii):-
[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]\frac{d y}{d x} = \frac{cos(t+sin t) (1+cost)}{cost}[/tex]
we know that
x = sin t
t = sin⁻¹ (x)
cost = √1 - sin²(t) = (√1-x²)
[tex]\frac{d y}{d x} = \frac{cos(x+sin^{-1}(x) (1+\sqrt{1-x^{2}) } }{\sqrt{1-x^{2} } }[/tex]
[tex]\frac{d y}{d x} = \frac{cos(x+sin^{-1}(x) (1+\sqrt{1-x^{2}) } }{\sqrt{1-x^{2} } }[/tex]
Put x =0 and y=0
[tex]\frac{d y}{d x} = 2[/tex]
d y = 2 d x
Integrating with respective to 'x' , we get
y(x) = 2 x + C
