Derivative of complex equation

In summary, the process for finding the derivative of an equation that requires both the chain rule and the product/quotient rule is to first apply the chain rule and then use the product/quotient rule on the resulting parts. The key is to take it step by step and not try to memorize formulas for every combination. Parentheses should also be used to avoid ambiguity.
  • #1
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hello!
1) what is the process to get the derivative of an equation that requires you to do first the chain rule and then the product/quotient rule, eg. sin(x^2(x+1))?
2) what is the process to get the derivative of an equation that requires you to do first the product/quotient rule and then the chain rule, eg. sin(e^x+1)(cos(x^2+1))?
thanks!
 
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  • #2
Step by step. Just do the first step first leaving d/dx symbols where they belong. Then see if more can be done to the resulting parts. There are too many combinations to try to memorize formulas for them all. Baby steps gets the job done.

I assume you mean (x^2)*(x+1) and not x^(2*(x+1)). Parentheses are your friends. Use them wherever it might otherwise be ambiguous.

d/dx( sin((x^2)(x+1)) ) = cos((x^2)(x+1)) * d/dx ((x^2)(x+1))
= cos((x^2)(x+1)) * [ d/dx x^2 ]*(x+1) + (x^2)[d/dx(x+1)])
= cos((x^2)(x+1)) * [ (2*x(x+1) + x^2) ]
 
  • #3
1. You just described it ... you do the chain rule first then the product/quotient rule.
2. Same here... you just described the process. It is the process of using the product rule followed by the pricess of doing the chain rule.

The trick is to do it in strps.

Eg 1.

##\frac{d}{dx}\left ( \sin\left ( x^2 (x+1) \right) \right)## ...put u=x^2 (x+1) and apply the chain rule.

##= \cos u \frac{du}{dx}## now evaluate du/dx using the product rule.

##\frac{du}{dx} = 2x(x+1) + x^2##

... now substitute back.
 
  • #4
Simon Bridge said:
The trick is to do it in strps.
stirrups? As in, astride a horse? :oldbiggrin:
 

Related to Derivative of complex equation

1. What is the definition of a derivative of a complex equation?

The derivative of a complex equation is a measure of how much the output of the equation changes with respect to the input. It is represented by the mathematical notation f'(z) or dy/dx, where z is a complex variable and x and y are real variables.

2. How is the derivative of a complex equation calculated?

The derivative of a complex equation can be calculated using the same rules as the derivative of a real function. However, since complex equations have both real and imaginary components, the derivative must be calculated separately for each component and then combined to form the complex derivative.

3. What is the significance of the derivative of a complex equation?

The derivative of a complex equation is important in many areas of mathematics and science. It allows us to analyze the rate of change of a complex function, find the slopes of curves in the complex plane, and solve optimization problems involving complex variables.

4. Can the derivative of a complex equation be complex?

Yes, the derivative of a complex equation can be complex. This is because the derivative is a measure of the change in both the real and imaginary components of the complex equation, so it can have both real and imaginary values.

5. How is the derivative of a complex equation used in real-world applications?

The derivative of a complex equation has many practical applications, such as in physics, engineering, and economics. It is used to analyze the behavior of complex systems, model the flow of fluids, and optimize the performance of complex systems. It is also used in the development of computer algorithms and in image processing.

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