Clear Filters
Clear Filters

How can I raise an anonymous expression to a power?

5 views (last 30 days)
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);
% The below line does not work, and I cannt figure out why...
quotient = (f1.*dg1dx - df1dx.*g1) / (g1).^2;
How can I use a power with this function?
  1 Comment
Dustin on 16 Jun 2014
Edited: Star Strider on 16 Jun 2014
Thank you for the answers. I got both of them to work for my problem. I am posting here because my next question si very related to my last. I am wondering how to know where to put the @(x) in anaonymous functions that later will take real-valued input. For example, I am trying to plot a function, its linearization and its 2nd order Taylor Polynomial on the same graph. I believe I can do this with ezplot correctly (though I cannot distinguish between the linearization and polynomial at all, so I could very well be wrong), however, I cannot get this to work with using "plot." My basic questions are: how do I know where to put the @ symbol? How does this affect whether or not the function takes real-valued inputs?
%%Taylor Polynomials
syms x
f = @(x) cos(x);
% dydx = diff(@(x) f(x), x);
dydx = @(x) diff(f(x), x);
dy2dx = @(x) diff(dydx(x), x);
a = 0;
% L(x) = f(a) + f'(a)(x-a)
Lx = @(x) f(a) + dydx.*(x-a);
Px = @(x) f(a) + dydx.*(x-a) + dy2dx.*(x-a).^2;
q = ezplot(f(x)); set(q,'Color','b'); hold on;
w = ezplot(Lx(x)); set(w,'Color','r'); hold on;
p = ezplot(Lx(x)); set(p,'Color','g'); hold on;
axis([-.05, .05, 0.95, 1.05]);
hold off; figure(2);
xnum = -5:0.1:5;
plot(xnum,f(xnum)); hold on;
%Neither of the below will work...
plot(x,dydx(x)); hold on;
plot(xnum,dydx(xnum)); hold on;

Sign in to comment.

Answers (2)

Rashmil Dahanayake
Rashmil Dahanayake on 9 Jun 2014
Edited: 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;

Star Strider
Star Strider on 9 Jun 2014
Edited: 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);
% 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)
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.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!