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Point) Differentiate f(x) = In Vx.f' (x)...

Question

Point) Differentiate f(x) = In Vx.f' (x)

point) Differentiate f(x) = In Vx. f' (x)



Answers

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\ {f(x)=x\left(x^{2}+3\right)} & {(2,14)} \end{array}$$

This question asked us to differentiate. The function after backs is X minus three. Sign of acts. Okay, We know the derivative of acts is simply the coefficient in front of the variable, which is one we know the derivative of negative three. Sign of axe is negative. Three. The cofounder stays the same. And then the derivative of sign we know is co sign. So we have one minus three.

All right. So we're driving the squared of X times sine of X. Which immediately I'm gonna write as X to the one half times sine of X. That's just looking towards us using the power rule. Now notice that we have A product rule here because we have something involving x times something involving X. one function here. The other function here. Yeah. So since we're doing the are the product rule, excuse me. When I derive Yeah, I must copy the original first and times it by the derivative of the second. Which in this case his co sign part of my s co sign of X. I'll say plus the original second function syntax times the derivative of the first one, which is one half X to the negative one half. Yeah. See you again. Yeah, highlight where each one is coming from now. All that's left to do is just clean up the problem. And really there's not much cleaning up to do. The book will probably right X to the 1/2 a square root of x. So I'll do that here and then on the right hand side, they'll probably move this X to the negative one half to the denominator, leaving us with something along the lines of Synnex over two square root of X. And this right here is your final answer with simplification.

The first thing that we could do is rewrite this function. Why? Basically rewriting square root of X is X to the power one half. It makes taking the derivative much easier. So then the left side and green is f on the right side is G and then we're going to use the product rule over here. So it just starts off with F, which is the original part of the left side times derivative of G, which is derivative of X to the one half That's one half X now to the negative, too. So we subtract one from that power. Now all of that plus, then we just do the opposite. So this time we take the derivative of F, which is derivative of X, is one, and then we leave g alone. So that's X to the one half. Okay, and then that is the derivative. You might see these switched, however, of course, because it's addition, then it doesn't matter what the order is, so they might be in a different order

This question asks us to differentiate the given function. We know that we can write this an exponent form to take away the radical. This will make it a lot easier to read now. Applying the chain rule, make the new expert or coefficient decreased export by one to get negative 4/5 and then take the using. The chan will take the derivative of the inside. The derivative of natural log of X is one over x. This simplifies toe act f prime of acts or the derivative is 1/5 x the fifth root of natural acts to the fourth. Remember, Negative exponents can be done as one over the positive exponent. So one over this value to the 4/5 can be written as this one.


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