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Chain Rule with Leibniz Notation

  • Thread starter opus
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  • #1
opus
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Homework Statement


Find the derivative of ##y=cos^3(πx)##
*Must be in Leibniz notation

Homework Equations




The Attempt at a Solution



(i) $$Let~ w=y^3 , y=cos(u), u=πx$$

(ii) $$\frac{dw}{dy} = 3y^2,~ \frac{dy}{du} = -sin(u),~ \frac{du}{dx}=π$$

(iii) By the Chain Rule,
$$\frac{dw}{dx} = \frac{dw}{dy}⋅\frac{dy}{du}⋅\frac{du}{dx}$$

(iv) $$= 3cos^2(πx)⋅-sin(πx)⋅π$$

(v) $$=-3πcos^2(πx)sin(πx)$$

I'm not sure if this is correct. We're told to use Leibniz notation, but were taught the Chain Rule in "prime notation", so I just put things together in what made some sort of sense. So I'd like to know if this is correct, and if I'm using this notation correctly.
Thank you for your time.
 

Answers and Replies

  • #2
Dick
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I looks just fine to me.
 
  • #3
opus
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Im surprised to hear that. Is there anything special about this notation? Why choose this over using primes? Or maybe its just a practice thing?
 
  • #4
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Im surprised to hear that. Is there anything special about this notation? Why choose this over using primes? Or maybe its just a practice thing?
The two notations aren't in a competition of better and worse. They both serve their own purposes. The Leibniz notation has the advantage that the variable is noted, which in case of more than one variable, or here, in case of change the variable makes sense. You cannot see this from the prime notation. Note that ##\dfrac{df(x)}{dx}=f'(x)## but ##\dfrac{df(x)}{dy}=0##. It also makes sense for future applications in differential geometry, as the ##dx## work as basis vectors there.

The prime notation is shorter and often does the job when it is clear what the variable, i.e. direction is, along which we differentiate. If there is only one, fine, but if there are more than one, it becomes necessary to distinguish them.

It also has another advantage which is usually hidden behind the notation. If you write ##f'(x)## then you normally mean the slope at ##x##. Let's call this point ##x=a##. Then what you really mean is ##f'(a)## which is ##\left. \dfrac{d}{dx}\right|_{x=a}f(x) = f'(a)##. However, we may also consider the function ##a \mapsto f'(a)## i.e. consider the dependency point → slope at this point. So the Leibniz notation forces you to be clear which role ##x## is actually playing.
 
  • #5
opus
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Ok that makes sense. I have to say that I didn't like it at first, but after finishing my homework I like it more for some of the same reasons that you stated. It seems more specific in that it specifies exactly what I'm working with. Thanks guys.
 

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