In Exercises 5-8, find ∇f at the given point. f(x, y, z)=(x^2+y^2+z^2)^-1 / 2+ln(x y z), (-1,2

step 1 The topic of this question is gradient vectors. In this question, we want to find the gradient v

In Exercises 5-8, find ∇f at the given point. f(x, y,...

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation involves taking the natural logarithm of a function before differentiating it. For functions expressed as products or powers, such as ln(xyz), logarithmic properties can simplify the differentiation process by converting multiplicative relationships into additive ones.

Chain Rule

The chain rule is a method for computing the derivative of a composite function. When a function is composed of another function, the differentiation process involves differentiating the outer function with respect to the inner function and then multiplying by the derivative of the inner function.

Gradient

The gradient of a function, denoted ?f, is a vector that consists of all of its first-order partial derivatives. It points in the direction of the greatest rate of increase of the function and its magnitude indicates the rate of increase in that direction.

Partial Derivatives

Partial derivatives measure how a function changes as each independent variable is varied while holding the other variables constant. They form the components of the gradient vector and are essential in multivariable calculus for analyzing functions of several variables.

Transcript

00:01 The topic of this question is gradient vectors.

00:07 In this question, we want to find the gradient vector of this particular function f at this particular point minus 1, 2, minus 2.

00:20 So first, let's find the gradient vector del f at the general point xyz in three -dimensional space.

00:32 Okay, so the gradient vector, remember, is defined as the vector, whose first component is the derivative of f with respect to x, the partial derivative, that is, and whose second component is the derivative with respect to y, and whose third component is the derivative with respect to z.

00:55 Okay, so let's find the first component.

00:59 We want to differentiate this whole thing with respect to x, and i'm going to start with the second term, which might be a bit easier.

01:13 Since we have a function of x inside another function, the long function, we have to use the chain rule.

01:23 And the way i like to think of that is differentiate the outside function with respect to the inside, treating x, y, z as a variable, and then multiply that by the derivative of the inside with respect to the variable you're differentiating with respect to your variable of interest.

01:53 So in our case, we're differentiating with respect to x, so the derivative is yz.

02:01 Notice that this just simplifies to 1 over x.

02:07 And so it's the same as if we had just a error.

02:11 Well, we can write this as a lot of x plus another term that doesn't depend on x.

02:18 And so that also tells us that the derivative is 1 over now, what's the derivative of this with respect to x? so this time we can't write it as a function of x plus a function that doesn't depend on x because we have this exponent on a sum.

02:41 So we have to use the chain rule.

02:45 So again, differentiate the outside with respect to the inside, and then multiply by the derivative of the inside with respect to x.

02:59 So there is our first component, the ihat component.

03:11 Now for the j hat component, we differentiate with respect to y.

03:18 So the derivative of this term will again be the derivative of the outside part, our function, with respect to x squared plus y squared plus that squared.

03:34 We already found that derivative up here.

03:38 So just use that.

03:40 And then we multiply that by the derivative of the inside, x squared plus y squared plus z squared, with respect to y, and that's 2y.

03:54 And similarly to before, for this term, we don't really have to use the chain rule.

04:00 We can write it as lon of y plus lon of x times z.

04:06 The derivative of this will be zero, and so we just have the derivative of a long y...

Please give Ace some feedback