22–36. Derivatives Find the derivatives of th...

Question

22–36. Derivatives Find the derivatives of the following functions.

f(x) = √coth 3x

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to find the derivative of the function F of X, which is equal to the square root of the hyperbolic cotangent of 5 X. So here we just need to find the derivative of our function F of X, and the first step we're going to do is actually rewrite our function F of X. So we're going to have F of X. equal to the quantity of the hyperbolic cotangent of 5 X. In quantity, raised to the one half power. So here we just rewritten the square root as raising that quantity to the 1/2 hour. That's gonna make it a little bit easier to take the derivative. So, we're looking for the derivative of F of X, so we're looking for F of X, and that's going to be equal to the derivative with respect to X. Of the quantity of the hyperbolic cotangent of 5 X. In quantity raised to the one half power. So applying our power rule. F is going to be equal to 1/2 multiplied by the quantity of the hyperbolic cotangent of 5 X. In quantity, raised to the minus 1/2 hour, but we need to also apply our chain rules, so we'll multiply this by the derivative with respect to X. Of the quantity of the hyperbolic cotangent. of 5 X. In quantity. So now we just need to take the derivative spec X of that hyperbolic cotangent function. That means that F of X is going to be equal to, first term remains unchanged, so this could be 1/2 multiplied by the quantity of the hyperbolic cotangent. Of 5X. In quantity raised to the minus 1 half power. And this is going to be multiplied by the quantity, and here we have to take the derivative spec X of the hyperbolic cotangent of 5 X. So we call that when you take the derivative of the hyperbolic cotangent function, you get minus the hyperbolic cosequence. Squared of the same argument, which our argument for the hyperbolic cotangent function was 5 X, so here we'll have 5x as the argument for the um hyperbolic cosequence squared function. And then again, we'll have to apply the chain rule, so we need to multiply that by the derivative with respect to X of the argument of our hyperbolic function, which is going to be 5 X. And so taking that last derivative, we see that F of X is going to be equal to one half, multiplied by the quantity of the hyperbolic cotangent of 5 X. In quantity raised to the minus 1/2 hour. Multiplied by the quantity of minus. The hyperbolic cosequent squared of 5 X. In quantity, and that gets multiplied by the derivative with respect to X of 5 X, which is just 5. So finally, we can just simplify our expression. So F of X is going to be equal to. I'll pull the min out front, so we'll have minus, and here we're going to have a fraction in the numerator, we'll have 5 multiplied by the hyperbolic cosequent squared of 5 X. And then that gets divided by the quantity of 2 multiplied by the square root. Of the hyperbolic cotangent. of 5 X. In quantity So, here we have just rewritten the quantity of the hyperbolic cotangent of 5 X in quantity rates to the minus 15, as the square root of the hyperbolic cotangent of 5 X, but in the denominator since we had a negative exponent. And the answer that we have here at the end is the answer that we're looking for for this problem. So, I hope that this explanation has been useful, and I'll see you in the next video.

22–36. Derivatives Find the derivatives of the following functions.

f(x) = √coth 3x

Key Concepts

Derivatives

Chain Rule

Hyperbolic Functions

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