Find the derivatives of the functions in Exercises 19–40.

Question

q = sin(t / (√t + 1))

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the derivative of P which is equal to cosine of the quantity of 2 X divided by the square root of the quantity of 2 X plus 1 in quantity. So here we need to find the derivative of P. So we're looking for P, and that's going to be equal to the derivative with respect to X of cosine of the quantity of 2 X divided by the square root of the quantity of 2 X plus 1 in quantity. And in order to take this derivative, we need to recall several of our derivative rules. So since we're taking a derivative cosine function, we need to recall that the derivative with respect to X of cosine of X. is equal to minus sign of X. We will also be taking the derivative of the square root function, so we have the derivative with respect to X. Of the square root of X. And recall that this is going to be equal to 1 divided by the quantity of 2 multiplied by the square root of X in quantity. We'll also need to use our constant rule for derivatives, so recall that the derivative with respect to X of a constant C is equal to 0. We'll also need our constant multiple rule. So recall that the derivative with respect to x of the quantity of C multiplied by X raises to the power n in quantity is equal to C multiplied by n, multiplied by X raised to the quantity of n minus 1. We're also going to need to use our chain rule to recall for the chain rule that we have the derivative with respect to X. Of F of G of X. And this is going to be equal to F of G of X. Multiplied by G of X. And then finally we'll be taking a derivative of a fraction, so we'll need to use our quotient rule. So recall for the quotient rule that we have the derivative with respect to X of the quantity of F of X. Divided by G of X. In quantity, and this is equal to the quantity of G of X. Multiplied by F of X minus F of X. Multiplied by G of X in quantity, divided by the quantity of G of X. In quantity squared. So we're gonna use all of these rules to calculate our derivative for P. And so P is going to be equal to, we'll first take a derivative of our cosine function. So we'll have minus sine. Of the quantity of 2 X divided by square root of the quantity of 2 X plus 1 in quantity. But we also need to apply our chain rule, so we need to multiply this by the derivative with respect to X. Of the quantity of 2 X. Divided by the square root of the quantity of 2 x plus 1 in quantity. So here we're going to be taking a derivative of a fraction. So we're gonna need to use our quotient rule. So it's advantageous to go ahead and define our F of X and G of X that appear up in our quotient rule. So, we'll set F of X equal to the numerator, which is equal to 2 X, and G of X is going to be our denominator, which is equal to the square root of the quantity of 2 X plus 1 in quantity. So when we calculate F of X, that's going to be equal to the derivative with respect to X of the quantity of 2 X. And so here we'll apply our constant multiple rule for derivatives. So this is going to be equal to our constant, which is 2, multiplied by the exponent, which is 1, multiplied by x raised to the quantity of 1 minus 1 in quantity. So simplifying this expression, this is going to be equal to 2. We also need to calculate G of X, and this is going to be equal to the derivative with respect to X of the square root of the quantity of 2 X plus 1 in quantity. And so here we're going to be taking the derivative of the square root function. So this is going to be equal to 1 divided by the quantity of 2 multiplied by the square root of the quantity of 2 X plus 1 in quantity. We also need to apply our chain rules, so this fraction needs to be multiplied by the derivative with respect to x of the quantity of 2 X plus 1. So this is going to be equal to 1 divided by the quantity of 2 multiplied by the square root of the quantity of 2 X plus 1 in quantity, and then that fraction is going to be multiplied by the quantity, and here we'll have the derivative with respect to X of 2 X. Again, we're going to use our constant multiple rule to take this derivative. So we'll have a constant, which is 2 multiplied by the exponent, which is 1, multiplied by x rates of the quantity of 1 minus 1. Then we'll have plus. The derivative of 1 with respect to X, and so here we'll use our constant rule, and so when we take the derivative of constant, that's going to be equal to 0. So simplifying this expression, this is going to be equal to 1 divided by the square root of the quantity of 2 X plus 1 in quantity. So now we have enough information to apply our quotient rule. So, coming back to P, we will have P is equal to, the first term remains the same, so we'll have minus sign. Of the quantity of 2 x divided by the square root of the quantity of 2 X plus 1 in quantity, and this gets multiplied by the quantity, and here we'll apply our quotient rule. So we'll have G of X, which is the square root of the quantity of 2 X plus 1 in quantity multiplied by F of X, which is 2. Then we'll have minus F of X, which is 2 X, multiplied by G of X, which is the quantity of 1 divided by the square root of the quantity of 2 X plus 1 in quantity. And that whole quantity is divided by the quantity of G of X squared. And so when a square G of X, I get 2 multiplied by X plus 1. So now we just need to simplify this expression. So, in the numerator of our fraction here, I can factor out a 1 divided by the square root of the quantity of 2 X plus 1 in quantity out of both terms and move that term into the denominator. So that means that P is going to be equal to minus sine. Of the quantity of 2 X divided by square root of the quantity of 2 X plus 1 in quantity, multiplied by the quantity. And here when I've factored out a one divided by the square root. Of the quantity of 2 X + 1, what I'm gonna be left with in the numerator is going to be. 2 X plus 1, that quantity multiplied by 2, and so I'll get 4 X plus 2, and then I'll have minus 2 X. And then that whole quantity is divided by the quantity of 2 X plus 1 in quantity multiplied by the square root of the quantity of 2 X plus 1 in quantity. Then I can simplify this expression a bit further, and I can factor out a factor of 2 out of the numerator and then collect like terms. So this is going to be equal to minus sine of the quantity of 2 X divided by square root of the quantity of 2 X plus 1 in quantity, and that is multiplied by the quantity of 2, multiplying the quantity, and here I'd have 4 X minus 2 X, which would give me 2 X, but I've factored out a 2, so this is gonna be X. Plus one, let's a factor out. A 2 from 2, and then that whole quantity is divided by the quantity of 2 X plus 1 in quantity, multiplied by the square root of the quantity of 2 X plus 1 in quantity. And finally, we're just going to rearrange the terms here, so that the sign term appears at the end. So this is going to be equal to minus the quantity of 2 multiplied by. X, the quantity of X + 1 in quantity. And that is divided by the quantity of 2 X plus 1 in quantity, multiplied by the square root of the quantity of 2 X plus 1 in quantity, multiplied by sine. Of the quantity of 2 X. Divided by square root of the quantity of 2 X plus 1 in quantity. And this here is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Find the derivatives of the functions in Exercises 19–40.

q = sin(t / (√t + 1))

Key Concepts

Chain Rule

Quotient Rule

Derivative of Trigonometric Functions

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