implicit Differentiation in MATLAB

what is the code with this matlab??

implicit Differentiation in MATLAB

Recall the steps for differentiating a relation y which implicitly depends on x:

1. Differentiate both sides of the equation with respect to x, remembering that y depends on x (so the derivative of every term of y is multiplied by dy/dx.
2. Solve the resulting equation for dy/dx.

Use the Solution Template to guide you through these steps to find dy/dx for the equation

cos( x/y ) = sqrt( x + y )

f=@(x) ; 
syms x;  % DO NOT CHANGE CODE ON THIS LINE
Lim_inf=limit( )
Lim_neginf=limit( )
p=fplot( );
axis([-50,50,-5,5]);  % DO NOT CHANGE CODE ON THIS LINE (reduces the y-axis to better see the asymptotes)

[Manish M]

Looks like a homework question. Can you please list down what all you've tried?

[Mona Mona]

i dont know how to solve this matlab at all...

Answer 1

[Jaynik]

Hi,

We can perform the implicit differentiation for the given equation using MATLAB as follows:

syms y;
f = @(x) cos(x/y) - sqrt(x + y); % Define the implicit function

% Differentiate both sides of the equation with respect to x as asked
dfdx = diff(f(x), x);
dfdy = diff(f(x), y);

% Solve for dy/dx
dy_dx = -dfdx / dfdy;

% Convert symbolic expression to function handle for plotting
f_dy_dx = matlabFunction(dy_dx, 'Vars', [x,y]);

% Calculate limits as x approaches infinity and negative infinity
% The limits will depend on the path y takes as x approaches infinity
Lim_inf = limit(dy_dx, x, inf);
Lim_neginf = limit(dy_dx, x, -inf);

% Plot the derivative dy/dx over a range of x values
% You need to specify a value for y when plotting
y_val = 1; % Sample value for y
p = fplot(@(x) f_dy_dx(x, y_val));
axis([-50,50,-5,5]);

Refer to the following documentation to learn more about these functions:

• diff: https://www.mathworks.com/help/symbolic/diff.html
• matlabFunction: https://www.mathworks.com/help/symbolic/sym.matlabfunction.html
• fplot: https://www.mathworks.com/help/symbolic/fplot.html

Hope this helps!