Answer:
Answer is [tex]f(1.01)=0.91[/tex]
Step-by-step explanation:
Since (1,1) lies on the curve it must satisfy it hence
[tex]x^{2}+sin(xy)+3y^{2}=C\\1+sin(1)+3=C\\\therefore C=4.017[/tex]
Now tangent line approximation of [tex]f(x)[/tex] is given by
[tex]f(x)=f(x_o)+f'(x_o)(x-x_o)[/tex]
Applying values we get
[tex]f(1.01)=f(1)+f'(1)(1.01-1)[/tex]
Now differentiating [tex]x^{2}+sin(xy)+3y^{2}=C[/tex] we get
[tex]2x+cos(xy)\times (y+x\frac{dy}{dx})+6y=0[/tex]
Solving for [tex]\frac{dy}{dx}[/tex] we get
[tex]\frac{dy}{dx}=\frac{1}{x}(\frac{-6y-2x}{cos(xy)}-y)[/tex]
Applying values we get
[tex]\frac{dy}{dx}=-9.00[/tex]
Using all the values we have obtained we get
[tex]f(1.01)=1-9.00\times (1.01-1)f(1.01)=0.91[/tex]