Use the definition of the derivative to determine d/dx (√ax+b)...

Question

Use the definition of the derivative to determine d/dx (√ax+b), where a and b are constants.

Accepted Answer

Welcome back everyone. Determine the derivative of the function G of X equals square root of C X plus d using the definition of the derivative where C and D are constants. A says C divided by 2 square root of C X plus D B D divided by 2 square roots of C X plus D. C. C divided by square root of C X + D and D D square roots. D divided by square root of C X plus D. So we want to identify the derivative of G of X, which is G of X. Let's write down the limit definition. We know that this is limit as H approaches 0 of the different quotient which is G of X plus H. Minus G of X divided by H. So all that we have to do is evaluate G at point X plus H, and we can do that by replacing X with X + H. So we get square root of C in C X plus H. Plus D. And now we're going to substitute this into the. Numerator, so we get limit as H approaches 0 of square root of. See X plus H + D And now we need to subtract the original function G of X, which is square root of C X plus D. All of that divided by H. So now, if we attempt direct substitution, well, we're going to get a form of 0 divided by 0, and because we are subtracting two radical terms. To eliminate a potential hole, we can just multiply both sides by the conjugate. And we're going to do that. So, let's multiply both sides by the conjugate. This means that we're multiplying both sides by square root of. See X plus H + D. So we're taking square root of that term and now instead of the negative sign, we're going to use the positive sign because this gives us the conjugate and rewrite the second radical term which is C X plus the square root of that, right? But because we're multiplying our Original fraction by the conjugate, we also have to divide by the conjugate to make sure that nothing really changes. We don't want to change the original function. And then we can apply the factorization formula for the difference of squares in reverse, right? So we get limit as each approach is 0. And now multiplying two conjugates, we have to square the first one, so we get C and C X plus H + D and subtract the square of the second term, so squaring square root gives us C X plus D. And because we are subtracting, we're subtracting C X and then subtracting D. Now in the denominator, we end up with H multiplied by. The conjugate. So we're going to add our conjugate. And then we're going to add res because we have a sum. And now let's notice that if we distribute the terms, we're going to get limit as H approaches 0 of CX minus CX, so those can be canceled out. Followed by CH. And the D values can be canceled out as well because we have D minus D. So we only have CH in the numerator. And if we have H, well, we can cancel it out because we also have H. In the denominator, so we only end up with C in the numerator. And for the denominator, we obtain the conjugate itself. So let's write down the conjugate once again for the denominator. What we're going to do from here is simply attempt direct substitution. We know that H approaches 0, so we get C in the numerator. Which essentially means that B and D, those should not be the correct answers, and if H approaches 0, then C multiplied by X plus H is simply C X. So we get square root of. C X + D plus another square root of CX plus D, which gives us two square roots of CX plus D. Which essentially means that the correct answer choice to this problem corresponds to a. Thank you for watching.

Use the definition of the derivative to determine d/dx (√ax+b), where a and b are constants.

Watch next