Asked by Dustin
on 9 Jun 2014

I am trying to replicate the quotient rule for finding derivatives using anonymous functions. I cannot figure out a way to raise one anonymous function to a power for this to work. Here is what I have

syms x

f1 = @(x) (x+1).*(x-3);

g1 = @(x) x.^3 + 2.*x +1;

h = @(x) f1(x)./g1(x);

df1dx = diff(@(x) f1(x), x);

dg1dx = diff(@(x) g1(x), x);

dhdx = diff(@(x) h(x), x);

simplify(dhdx)

% The below line does not work, and I cannt figure out why...

quotient = (f1.*dg1dx - df1dx.*g1) / (g1).^2;

simplify(quotient)

How can I use a power with this function?

Answer by Rashmil Dahanayake
on 9 Jun 2014

Edited by Rashmil Dahanayake
on 9 Jun 2014

Firstly correct the definition the inline functions for differentiation. Start the command with @ symbol. When calling inline functions you must pass an argument. eg F(x).

Updated code

syms x

f1 = @(x) (x+1).*(x-3);

g1 = @(x) x.^3 + 2.*x +1;

h = @(x) f1(x)./g1(x);

df1dx = @(x) diff(f1(x), x); % note the location of @ symbol

dg1dx =@(x) diff(g1(x), x);

dhdx = @(x) diff(h(x), x);

simplify(dhdx(x)) % pass an argument to the function

quotient = (f1(x).*dg1dx(x) - df1dx(x).*g1(x)) / (g1(x)).^2;

simplify(quotient)

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Answer by Star Strider
on 9 Jun 2014

Edited by Star Strider
on 9 Jun 2014

The Symbolic Toolbox will work with anonymous functions, but it prefers not to, because anonymous functions prefer numeric inputs.

‘Undefined function 'power' for input arguments of type 'function_handle'.’

Try this instead:

syms x

f1(x) = (x+1).*(x-3);

g1(x) = x.^3 + 2.*x +1;

h(x) = f1(x)./g1(x);

df1dx = diff(f1(x), x);

dg1dx = diff(g1(x), x);

dhdx = diff(h(x), x);

simplify(dhdx)

% The below line does not work, and I cannt figure out why...

% (It does now!)

quotient = (f1.*dg1dx - df1dx.*g1) / (g1).^2;

quotient = simplify(collect(quotient))

% Compare:

quotient2 = diff(f1/g1, x);

quotient2 = simplify(collect(quotient2))

Not surprisingly, the same answer.

If you want to use anonymous functions outside of the Symbolic Math Toolbox to calculate numeric derivatives, it‘s easy enough:

dfdx = @(f,x) (f(x+1E-8)-f(x)) ./ 1E-8; % Generic derivative function

f = @(x) x.^2; % Function to differentiate

x = 0:5; % Evaluate function & derivative

fx = f(x)

dfx = dfdx(f,x)

produces:

fx =

0.0000e+000 1.0000e+000 4.0000e+000 9.0000e+000 16.0000e+000 25.0000e+000

dfx =

10.0000e-009 2.0000e+000 4.0000e+000 6.0000e+000 8.0000e+000 10.0000e+000

With your functions:

f1 = @(x) (x+1).*(x-3);

g1 = @(x) x.^3 + 2.*x +1;

quotnt = f1(x)./g1(x)

dquotnt1 = (f1(x).*dfdx(g1,x) - dfdx(f1,x).*g1(x)) ./ (g1(x)).^2

dquotnt2 = dfdx(@(x) f1(x)./g1(x),x)

The result produced by dquotnt1 is negative. The dquotnt2 expression gives the correct result.

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## Dustin (view profile)

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