9–61. Evaluate and simplify y'.

y = x^√x...

Question

9–61. Evaluate and simplify y'.

y = x^√x+1

Accepted Answer

Welcome back, everyone. Find the derivative of the function Y equals 2 x to the power of 3 square root of x minus 2. Let's notice that in this problem our function Y is F of X, raise to the power of another function G of X, right, and essentially to differentiate it successfully we're going to apply logarithmic differentiation. So we're going to take LN of both sides, which gives us LNN of Y equals LN of. 2 Xs raises the power of 3 square roots of X minus 2. And this is really helpful because now we can rewrite. LN of Y equals and let's apply the power rule. The power rule says that we can take the exponent and bring it down in front of the log. So it gives us LN of Y equals 3 square root of x minus 2 multiplied by LN of 2 X. Moreover, we can say that LN of Y is equal to. 3 square root of x minus 2 multiplied by LN of 2 plus LN of X, right, because we have a product of 2 multiplied by X, and based on the product rule we can write this as a sum of two logs, a line of 2 and a line of X. And from here we're going to differentiate both sides. So let's begin. Let's recall that the derivative of LN of X is 1 divided by X. So in this case on the left hand side we're going to get 1 divided by Y, but Y is a function of X, and this means that we want to apply the chain rules so we're multiplying by Y and now we have to focus on the right hand side. And here we have a product of. Two functions, right? So essentially what we want to do is apply the product rule. Let F. Be equal to our first function which is 3 square root of. X minus 2. And then our second function is going to be G equals LN 2 plus LN X. And based on the product rule we have to recall that first rule we're taking the derivative of the first function. We can factor out the constant and differentiate 3 multiplied by x minus 2, raise the power of 1/2, so we are rewriting square root in terms of exponent. To successfully differentiate it, right? So we're taking the derivative of F, multiplying that by the second function which is G, so we're multiplying that by L and of. 2 X, let's combine those terms, right? It will not make differentiation easier, so we can just write that we get. We multiplied by the derivative of X minus 2 to the power of 1/2 multiplied by LN of 2 X, and then based on the product rule we are adding the first function which is 3 square root of. X minus 2 multiplied by the derivative of G, right? So in this case, that would be the derivative of LN2 plus the derivative of LN of X. So now what we're going to do is simply rewrite it. We want to solve for Y, so Y equals. We have to multiply both sides by Y, so we're going to get Y in front, and then we want to evaluate the derivative, right? So based on the power rule for x minus 2 raise the power of 1/2, we're going to bring down the exponent, and this gives us we multiply it by 1/2 multiplied by x minus 2, raise the power of -12, and based on the chain rule, we have to multiply by the derivative of the inner function which is X minus 2. And then we also have Alan of 2 Xs. Then we are adding 3 square roots of. X minus 2 multiplied by the derivative of LN2, which is 0 because it is a constant so we can immediately say that it's 0, and then the derivative of LN X is 1. Divided by X. And that's it. We also want to recall that the derivative of X minus 2 is simply 1 based on the power rule because the derivative of X is 1, the derivative of -2 is 0. It is a constant. So what we want to do is simply rewrite Y as Y to begin with, so we're going to substitute Y as 2 X raises the power of 3 square root of x minus 2. And then in parties, we just want to make sure that it looks better. So we have 3. Multiplied by LN of 2 X. Divided by the reason why we are dividing is because we have 1/2 to begin with. We're going to write 2 and then we have a negative exponent. So x minus 2 to the power of -12 is the same as 1 divided by square root of x minus 2. Right, so we're using the rules of exponents and the fact that it is a negative exponent so we can just rewrite that negative exponent as a positive exponent in the denominator, and an exponent of 1.5 represents square root. So this, this is how we get our first term and then we're going to add. 3 square root of X minus 2 divided by X because we're multiplying by 1 divided by X. And and that's it. We have our final answer. The derivative of Y, which is Y is equal to 2 X to the power of 3 square root of X minus 2 multiplied by 3 LN of 2 X divided by 2 square roots of X minus 2 plus. 3 square root of X minus 2 divided by X. Thank you for watching.

9–61. Evaluate and simplify y'.y = x^√x+1

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9–61. Evaluate and simplify y'.

y = x^√x+1