49–55. Derivatives of tower functions (or g^h) Find the deriva...

Question

49–55. Derivatives of tower functions (or g^h) Find the derivative of each function and evaluate the derivative at the given value of a.

h (x) = x^√x; a = 4

Accepted Answer

Determine the derivative of F of X equals 3X, raise the square root of 2 X, and evaluate F of 2. Now, to solve this, we are first going to write this equation as Y equals 3 X, raised to the square root of 2 X. And then take natural log of both sides. We get natural log of Y equals natural log, 3 X raise the square root of 2 X. Now, the reason we did this is to make use of a log property. Which you look at this first. We have natural log of 3. Multiplied by X, rates to the square root of 2 X. Giving us natural log of 3 plus natural log. Of X-rays to the square root of 2 X. This is our private role for natural logarithm. Which tells us if we have natural lag. Of A multiplied by B. It can be rewritten as natural log of A plus natural log of B. Now from here, We can make use of another log property. Which tells us that if we have natural lag, Of a raised to the B, it can be be written as B natural log of A. So, we now have natural log of Y equals natural log of 3. Plus the square root of 2 X, natural log of X. Now, we can take the derivative of our equation. We'll take the derivative of both sides. And for the left side, we have the derivative of natural log of Y, which ends up being one divided by Y. Why? Because we are in terms of Y for a derivative. In our functions in terms of X, we will have to add this Y. The derivative of natural like of 3 ends up being zero. And that derivative on the right here ends up being a product rule. Now the private rule Is given by U multiplied by V. Plus you multiplied by V. Where our function is a product of two different functions, U and V. In this case, U is the square root of 2 X, V is natural log of X. Meaning we can find our derivatives U and V. By taking a couple rules. Now, the square root of 2 X is actually a small chain rule problem. Where the square root of 2 X's derivative is 1 divided by 2 square root of 2 X. Now, that's just the outside function. We also multiply by the derivative of the inside function, which is 2. For natural log of X, this derivative is a known derivative being one divided by X. So with our product pool, we will have You Which is 1 divided by 2 squares of 2 X multiplied by 2. All of this multiplied by V. Which is natural log of. Plus U squad of 2 x multiplied by the derivative. Of V1 divided by X. We can now simplify To get one Divided by a square of 2 X, multiplied by natural log of x. Plus square of 2 x divided by X. Giving us 1 divided by Y, Y equals natural log X divided by square root of 2 X. Plus we're of 2 x divided by X. Now, We can actually combine these into a single fraction. By getting the same denominator. We will multiply The right term By square of 2 X divided by square of 2 X, and the left term. By X divided by X. We end up getting one available by way by prime. Equals X, natural log of X. Plus 2 X all divided by X square root of 2 X. Now, we want Y by itself, so let's Multiply both sides. By why We didn't end up getting Y is equals to X natural log X plus 2 X divided by X2 root of 2 X. All multiplied by y. And now We can do a substitution or why. We said Y was equals to 3 X raised to the square root of 2 X. Which means we can rewrite our function. As Y equals X, natural log X plus 2X divided by X square root 2 X. All multiplied by 3 X rated at the square root of 2 X. No. We can go ahead and plug in our value. But first, just quickly notice that there's X on all of our terms in the numerator, and an X in our term on the denominator, which means they can be canceled out. We have Y equals natural log X plus 2. Divided by a square root of 2 X, multiplied by 3 X. Raised to the square root of 2 X power. The problem told us then. That we want to plug in the value F. Of 2, or rather we want to solve for F 2 by plugging in X equals 2. So we will say then. F of 2 is equals to natural log 2 + 2 divided by the square root of 2 multiplied by 2. All multiplied by 3. Multiplied by 2, raise the square root of 2 multiplied by 2. Now let's simplify all of this. Natural lag 2 + 2 on the top, divided by the square root of 4 on the bottom. Multiplied by 3 Multiplied by. 2 raises to the square root of 4 power. This gives us natural log 2 + 2 divided by 2. Multiplied By 3 And 2 squared. We end up getting 3 multiplied by 2 squared, which is 4. That goes to 12, multiplied by natural log of 2, plus 2. All divided by 2. This gives us 6 multiplied by natural log of 2 + 2. And finally, we can distribute our 6 to get 6, natural log of 2, plus 12. This is our simplified answer, which means this is our final answer being 6 natural log 2 + 12. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

49–55. Derivatives of tower functions (or g^h) Find the derivative of each function and evaluate the derivative at the given value of a.h (x) = x^√x; a = 4

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49–55. Derivatives of tower functions (or g^h) Find the derivative of each function and evaluate the derivative at the given value of a.
h (x) = x^√x; a = 4