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Find the differential of each of the given functions.$$y=6 x \sqrt{1-4 x}$$

$d y=\frac{6(1-6 x)}{\sqrt{1-4 x}} d x$

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 8

Differentials and Linear Approximations

Derivatives

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seven This problem. We've been given a differential equation, which I've written in the top left. Now, this is a Benue equation where n equals 1/2 p of X is two times x a pair of minus one cure fax is six times the square root of one plus X squared. So as it's a bit nearly equation, we can make the substitution if you equals y to the power of one minus and one minus end is just one minus 1/2. So this is why the power of 1/2. So using equation 1.8 point 12 from the textbook, this transforms the differential equation. So it looks like one over 1/2 the utx plus two x to the minus one times white and 1/2 equal to six times one plus six times square root of one plus X squared. Now, if we multiply fruit by ah half, we get that this equation is equal to D u D X plus X to the minus one. Why spare? Half is equal to free times Square root of one plus X squared of this differential equation could be solved using an integrating factor. Now the integrating factor we propose is off the exponential of the integral one over X, which is excellent minus one The X now the Inter court one over axes just for natural log of the absolute value of X. This gives is eat about natural log the absolute value of X, the exponential in the natural World Council. So we get that the integrating factor is the absolute value of X From the question we know that X is always positive. Service is simply X. We use this. This means the differential equation can be reversing is the differential of you X with respect x and that is equal to free X I'm square root of one plus X squared. Now we want to integrate both sides to get our solution. This means we need to integrate free x times the square root of one plus x squared with respect to X. We do this by making the substitution off. So we make the substitution off T equals one plus X squared and this causes DT to be equal to two x the X make the substitution into the integral. This interval becomes the integral off free over two. I was square root of tea. E T. Andi has solution t to the power of free over two, which is, you know, original X variables, which is equal to one plus X squared times to the power of free over two, plus some integration constant C. So if we go back to the solution of this differential equation, this means. But we have you Times X is equal to want plus X squared about free over two. Let's see we divide through by X. This gives us that the right hand side is equal to one over X times one plus x squared to the power of free over two plus C over x onda. We remember that you is equal to y to the power of half. So our solution in terms of why is that? Why, to the power of 1/2 is equal to one over x times one plus X squared forever tip for C over X

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