Sal differentiates f(x)=-4∛x, and evaluates the derivative at x=8. This can actually be done quite easily using the Power rule!
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- Why could we have not just cubed -4 and cube root of X? I thought it would work, but I got -64.(2 votes)
- I guess I got your mistake. I guess you did this:
y = (-4) * (cuberoot of x) || raise every term to the power of 3
=> y^3 = (-4^3) * (x)
= y^3 = -64 * (x)
And here is your problem: you finally erased the cube root, but now you have y raised to the third power on the left hand side. It would be incorrect to simply work with y^3 here. I guess you forgot that you raised both sides of the equation.
But it's fine to make mistakes like these, because it shapes your skills.(3 votes)
- I still don't understand how he differentiated a radical(2 votes)
- If you're not quite understanding the relationship between radicals and exponents, you may want to review this video: https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/rational-exponents-intro/v/basic-fractional-exponents(2 votes)
- When Sal moved the -2 exponent from -2/3 to get 1/3. I guess it's because looking at the function, it would be much easier to simplify 8^1/3 than 8^2/3
? But, if he solved 8^2/3, he would still get 4, right?(1 vote)
- [Voiceover] Let's say that we have a function, f of x, and it is equal to negative four times the cube root times the cube root of x. And what we want to do is we want to evaluate the derivative of our function when x is equal to eight. So see if you can figure this out. All right, now this might look foreign to you. You might say, well, I've never taken a derivative of a cube root before, but as we'll see, we can actually just apply the power rule here. Because this function can be rewritten as, so f of x can be rewritten as negative four. The cube root of x is the same thing as x to the one-third power. And now it might be a little bit clearer that we can apply the power rule. We could take this one-third, multiply it by this co-efficient, negative four. So we have negative four times one-third, and you have times x to the one-third. We just decrement that exponent to, that's a different shade of blue, to the one-third minus one power. This is the derivative. So f prime of x is equal to that, and so now we just have to simplify. This is equal negative four-thirds times x. One-third minus one is negative two-thirds power. And so if we want to evaluate f prime of eight, f prime of eight is equal to negative four-thirds times eight to the negative two-thirds power. Well that's the same thing as negative four-thirds times eight, eight to the one-third, and then raise that to negative two power. I'm just using exponent properties here. If this looks completely unfamiliar how I got from that to that, I encourage you to review exponent properties on Khan Academy. Well eight to the one-third, that is just two. So this is just two, and then two to the negative two power. Remember, let me just take some steps here. It's a good review. This is equal to negative four-thirds times two to the negative two is the same thing as one over two to the two. These two things are equivalent. That and that. One over two to the two is the same thing as two to the negative two. So this is one over four, and so this is going to simplify to negative one over three. Is equal to negative one over three, and we are done.