Show that the operation of taking the gradient of a function has the given property. Assume that

Show that the operation of taking the gradient of a function...

Key Concepts

Gradient

The gradient is a vector operator that assigns to each point the vector of partial derivatives of a scalar function. It points in the direction of the steepest ascent of the function and its magnitude represents the rate of increase, forming a fundamental tool in multivariable calculus.

Linearity of the Gradient

The linearity property of the gradient states that it distributes over addition and scalar multiplication; that is, the gradient of a sum of functions is the sum of their gradients and constants can be factored out. This property follows from the linearity of differentiation and is essential in simplifying complex differential expressions.

Product Rule for the Gradient

When taking the gradient of the product of two differentiable functions, the product rule applies. It states that the gradient of the product is equal to one function times the gradient of the other plus the other function times the gradient of the first. This rule parallels the product rule in single-variable calculus and is key in handling products within multivariate differentiation.

Quotient Rule for the Gradient

The quotient rule for the gradient extends the concept of the single-variable quotient rule to functions of several variables. It provides a method for differentiating a ratio of two differentiable functions by combining the gradient of the numerator and denominator in a structured way, crucial when the function is expressed as a quotient.

Chain Rule

The chain rule is a method for differentiating composite functions. In the context of the gradient, it allows one to differentiate a function raised to a power by taking the derivative of the outer function evaluated at the inner function and then multiplying by the gradient of the inner function. This concept is indispensable in handling compositions of functions in multivariable calculus.

Transcript

00:01 In this problem, we are asked to prove the four statements that they give us.

00:04 And so our first statement is that the gradient vector of a times a function u plus b times a function v is equal to a times the gradient vector of u plus b times the gradient vector of v.

00:20 And so to prove this, we know that to get the gradient vector of any function in general, we're going to take the partial with respect to x.

00:36 Of that function right here and that's going to be our first component of the vector and our second component is going to be the partial with respect to y of that function and now we're going to take the partials of the functions or the single function i should say and when we do that we get a as a constant so a can come out front and so we get a times the part of the part of partial of u with respect to x plus b which is also a constant times the partial of b with respect to x and then we're going to get again a out front partial of u with respect to y plus b times the partial of v with respect to y and now we can take an a out of the components of these two parts of both of the components and a b out of these two parts.

02:16 And when we do that, we get the following.

02:22 We're going to multiply a times the vector, partial of u with respect to x, comma, partial of u, times the partial of v with respect to x and the partial of b with respect to y.

02:55 Now, we know that this vector here makes up the gradient vector for u because we have the partial with respect to x and with respect to y.

03:06 And so this is equal to a times the gradient vector for you.

03:11 And we know that this here is equal to the gradient vector for v because we have the partial with respect to x and with respect to y and so we add to this b times the gradient vector for v and we have proven what they gave us and so now we're going to move on to part b in which they ask us to prove that the gradient vector for some function uv is equal to u times the gradient vector for v plus v times the gradient vector for u now to prove this first we are going to write out what we know the gradient vector would be just taking the gradient vector of this function and so when we do that we would get with the gradient vector for uv is equal to the partial of uv with respect to x this is a vector again and then the the second component is going to be the partial of uv with respect to y.

04:47 And now we are going to find these different partials by using the product and chain rule together.

04:59 And so we know that the partial of uv with respect to x is equal to v times the derivative of x, which is the partial of u with respect to x plus u times the partial of v with respect to x and now we want the partial or now we want to find so that was for this partial right here and now we want to find this one and so we know that the partial of uv with respect to y is going to be equal to v times the partial of u with respect to y plus u times the partial of v with respect to y.

06:08 And again, we got both of these by using the product and chain rule.

06:13 And so knowing this, we can then plug those back in to our vector up here.

06:19 And when we do that, we get that the gradient vector for uv is equal to v times the partial of u with respect to x plus u times the partial of v with respect to x comma v the partial of u the partial of u with respect to y plus u times the partial of v with respect to y and we can pull out a v from both components and a u from both components to get two different vectors and so when we pull out the vs from these parts of the components, we get v times the vector, partial of u with respect to x, comma partial of you with respect to y.

07:23 And we're going to add to that a vector that we pull u out of.

07:28 And so we add to that u times the vector, partial of v with respect to x, partial of v with respect to y.

07:45 And we know that this is the gradient vector for u, and this is the gradient vector for v.

07:55 And so therefore, we have shown that the gradient vector for uv is equal to v times the gradient vector for uv is equal to v times the gradient.

08:06 Vector for u plus u times the gradient vector for v and you can easily switch the order of these but we have proven what they have asked us to okay and now our third thing to prove is that the gradient vector of some function u over some function v is equal to v times the gradient vector for u minus u times some gradient vector for v over v squared.

09:05 Okay, so our first step in proving this is going to be to write out what we know the gradient vector for u over v is going to look like.

09:20 And so that is that the gradient vector for u over v is equal to the partial with respect to x of u over v again this is a vector because we're doing it we're taking the gradient vector of u over b then comma the partial with respect to y of u over v now we're going to do the following we're going to use the chain rule and the quotient rule to determine what these partials are and so we know that the partial with respect to x of u over v is equal to the denominator which is v times the derivative of the top which is going to be the partial of you with respect to x and then we're going to subtract the denominator or excuse me the derivative of the denominator which is the partial of v with respect to x, and we're going to multiply that times the numerator, which is u...

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