First and second derivatives Find f′(x),f′′(x).
f(x) = x...

Question

First and second derivatives Find f′(x),f′′(x).

f(x) = x/x+2

Accepted Answer

Welcome back, everyone. In this problem, given the function H X equals X divided by X minus 5, determine the first and the second derivatives of H of X. A says the first derivative is -5 divided by X minus 5 squad, and the second derivative is 10 divided by X minus 5 to the 4th. B says they are 5 divided by X minus 5 squared and the 10 divided by X minus 5 cubed respectively. C-5 divided by x minus 5 squared and 10 divided by X minus 5 cubed and D-5 divided by X minus 5 squared and 10 divided by X minus 5 cubed. Now, if we're going to find the 1st and 2nd derivatives of HF X, we first need to ask ourselves, how can we differentiate H of X, which is a constant? Well, we can use the quotient rule. Recall, OK. That for H X equals a quotient, let's say the quotient is u divided by V, then the derivative of H of X is going to be equal to U minusuv divided by Vquad where U and V are functions of X. In other words, if we can differentiate U and V, then we can plug them along with the values of U and V into the quotient formula, or sorry, into the quotient rule of differentiation to differentiate H of X. And then to find the second derivative of H of X, we would just differentiate the first derivative again. So let's start by finding the first derivative of H of X, OK? Now here, OK, uh let me, let me write this in blue here. So first derivative, I'm gonna put that in blue and the second derivative in red. So starting with the first derivative, let's let you, OK. I forgot my color for a minute. Let's let you be equal to our numerator X, OK. And we, as you can see from our from our quotient rule here would be the denominator, X minus 5. Well, if U is X, then the derivative of U is going to be 1 because the derivative of X is 1, and the derivative of V equals 1 as well. Because when we differentiate negative 5, it would be 0 because it's a constant. Know that we have the derivatives of U and V. By applying the quotient rule, that means the first derivative of H of X is going to be equal to. U 1 multiplied by V X minus 5. Minus U, which is X, multiplied by V which is 1. And now all of that is being divided by V X minus 5 all squared. If we expand our brackets here, this is going to be X minus 5 minus X. A divided by x minus 5 r 2. And no X minus X is 0, so the first derivative is going to be equal to -5 divided by X minus 5 squared. So now we have our first derivative of H. How can we find the second derivative? Well, like we said earlier, we can differentiate the first derivative, OK? But no, when differentiating this derivative, when differentiating the derivative of H of X will have different values for U and V. So applying the quotient rule, applying the quotient rule to the second derivative. OK. No, OK. We're gonna have our numerator as -5, so U is -5 and V is going to be X minus 5 squared. OK? Now, if we think about it here. OK, then if we let U be equal to -5, then the derivative of U equals 0 because -5 is a constant, and if V equals X minus 5 or squad, then we can differentiate V using the chain rule means that we'll bring down the power 2 multiply by the derivative of inside the bracket 1 because the derivative of X minus 5 is 1. And then subtract 1 from our term. So that's X minus 5 raises to the power of 2 minus 12 multiplied by 1 is 22 minus 1 is 1, so the derivative of V is 2 multiplied by X minus 5. Know that we have the derivatives of U and V, OK then. That means We can find the second derivative, OK, of H of X, which is now gonna be, like we said, U 0 multiplied by V X minus 5 square minus V2 multiplied by X minus 5 multiplied by U-5. I know all of this is going to be divided by V squared. So that's X minus 5 squared. Squeered Now, when we simplify this here, is 0 multiplied by X minus 5 squared is 0. If we expand our brackets here, that's gonna be uh -2 multiplied by X minus 5, OK. So this is gonna be negative 10 X. OK. Plus 10, sorry, -2 X, my apologies. -2 X plus 10. That's gonna be multiplied by -5. And again, all of this is being divided by X minus 5 to the 4th. And no, we simplify a numerator even further, -2 X multiplied by -5 is. -2X multiplied by -5 is 10 X, OK? And 10 multiplied by -5 is -50. It's divided by X minus 5 to the 4th. And now we have a unique situation here because no, we can cancel, we can rewrite or enumerate as 10 multiplied by X minus 5. And now we can cancel X minus 5 from our numerator and take 1 from our denominator. So that's gonna leave us with 10 divided by X minus 5 cubed. Thus, the second derivative of H of X. It's going to be 10 divided by X minus 5 cubed. If we look back at our answer choices, then for our values of the 1st and 2nd derivatives of H of X, then C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

First and second derivatives Find f′(x),f′′(x).
f(x) = x/x+2

Key Concepts

First Derivative

Second Derivative

Quotient Rule

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