implicit Differentiation in MATLAB

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Mona Mona
Mona Mona 2019-6-23
编辑： Jaynik 2024-6-20
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)
  2 个评论
Manish M
Manish M 2019-6-23
Looks like a homework question. Can you please list down what all you've tried?
Mona Mona
Mona Mona 2019-6-23
编辑：Mona Mona 2019-6-23
i dont know how to solve this matlab at all...

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回答（1 个）

Jaynik
Jaynik 2024-6-20
编辑：Jaynik 2024-6-20
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:
Hope this helps!

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