Use logarithmic differentiation to find the derivative of the given function.y=(x+2x)23(x2−4)y=\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}y=(xx+2)32(x2−4)

[ 1 3 ( x + 2 ) − 1 3 x + x x 2 − 4 ] ⋅ ( x + 2 x ) 2 3 ( x 2 − 4 ) \left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} [ 3 ( x + 2 ) 1 − 3 x 1 + x 2 − 4 x ] ⋅ ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

Use logarithmic differentiation to find the derivative of the given function. y = ( x + 2 x ) 2 3 ( x 2 − 4 ) y=\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} y = ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90l0 -0c4,-6.7,10,-10,18,-10 H400000v40H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5c53.7,-170.3,84.5,-266.8,92.5,-289.5zM1001 80h400000v40h-400000z">

Here, we're asked to use logarithmic differentiation in order to find the derivative of the given function. Here, we're given that y is equal to the square root of x plus 2 over x to the power of 2 thirds times x squared minus 4. Now since all of this is under a square root and I have functions multiplying each other with fraction exponents, I know that logarithmic differentiation is going to make this process easier. So it's not that we have to use log differentiation to find this derivative. It's that it's gonna be easier. So let's get into things here. Now because this is under a square root, remember that we can also think of this as being raised to the power of 1 half. So if I rewrite this function, this is x+2 over x to the power of 2 thirds times x squared minus 4, all of this is raised to the power of 1 half. That's gonna make it easier to use our log properties, having that already written as a power. So here we have the natural log of y is equal to the natural log of this mess over here. This is how we start with our log differentiation. Now we wanna use our properties of logs in order to rewrite this and expand it out. So that natural log of y is going to stay as is. There's nothing I need to expand out there. But over on this side, I know the first thing that I can do do is pull that one half out front. So this is 1 half times the natural log of x plus 2 over x to the power of 2 thirds times x squared minus 4. So what can I do further to expand this out? Because I this doesn't look super easy to deal with yet. But since these are multiplying each other, I can go ahead and separate them out as adding. Remember, this one half is going to be multiplying everything going on in here, so I wanna make sure that I keep that organized. So this is the natural log of x plus 2 over x to the power of 2 thirds. Note that we're taking the natural log of all of that. And then plus the natural log of x squared minus 4. So this one here, this natural log here looks pretty simple, but this one still has an exponent. And I can go ahead and pull that exponent out front. So this then gives me 1 half times the natural log times 2 thirds, the natural log of x plus 2 over x plus the natural log of x squared minus 4. Now this does look a bit easier to deal with, but I can take this even further. Because I have these two functions being divided, I can split this out into 2 natural logs being subtracted. We wanna make sure to keep all of our constants organized here, and we know what is multiplying what. So this is then 2 thirds times the natural log of x plus 2 minus the natural log of x. Then this is plus, so that's in parentheses being multiplied by that 2 thirds, plus the natural log of x squared minus 4. Okay. So I think we've expanded this out in as many logs as we can, but let's go ahead and rewrite this a little bit and get all of our constants sorted out here. Because this one half gets distributed to absolutely everything in these parentheses, and this 2 thirds gets distributed to these 2. I need to think about exactly what's going on here. So this first term is 1 half times 2 thirds times the natural log of x plus 2. This simplifies to just a 1 third natural log of x plus 2. And then I'm subtracting here. This 2 thirds also is on that term as is the 1 half. So this is another 1 third. This is 1 third times the natural log of x. And then plus, this term only has the one half multiplying it. So this is 1 half times the natural log of x squared minus 4. And then on this side, I still just have that natural log of y, just carrying that all the way down. Now from here, we can actually take the derivative. So I wanna take ddx on both sides. I've done everything I can do with my log properties, and now I just need to take the derivative. Now when taking the derivative with respect to x of the natural log of y, this will give me one over y dydx using the chain rule as we have before. And over here on this side, the derivative of a 1 third natural log of x plus 2 is going to give me a 1 third times 1 over x plus 2. Now the derivative of x plus 2 is just 1, so that's not going to change anything there when I think of the chain rule. Then over here, I have a 1 over 3 times 1 over x, because that's just the natural log of x. That derivative is just 1 over x. And then finally adding this together, 1 over 2 times 1 over x squared minus 4. And I do need to apply the chain rule here because the derivative of x squared minus 4 is 2 x. Now we can do some simplification here, and we can also pull this y over multiplying by it to give us that dydx that we want. So dydx is then equal to all of this stuff over here. Now doing some simplification, this 2 cancels with that 2, and I can combine everything all together, or I can condense those fractions a little bit. This gives me 1 over 3 times x+2minusone over 3x plus x over x squared minus 4. So that's everything that's inside of my brackets there. And then I'm multiplying this by my original function y, which we can go back up here to take a look at what that original function is. It is this mess under my square root. So this all gets multiplied by the square root of x+2overxtothepowerof2 thirds times x squared minus 4. And this is my whole derivative, and I didn't have to use the chain rule, the quotient rule, the product rule altogether. We could just use a log differentiation. Feel free to let us know if you have any questions, and I'll see you in the next one.

Use logarithmic differentiation to find the derivative of the given function. y = ( x + 2 x ) 2 3 ( x 2 − 4 ) y=\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} y = ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

[ 1 3 ( x + 2 ) − 1 3 x + x x 2 − 4 ] ⋅ ( x + 2 x ) 2 3 ( x 2 − 4 ) \left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} [ 3 ( x + 2 ) 1 − 3 x 1 + x 2 − 4 x ] ⋅ ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

1 3 ( x + 2 ) − 1 3 x + x x 2 − 4 \frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4} 3 ( x + 2 ) 1 − 3 x 1 + x 2 − 4 x

[ 2 3 ( x + 2 ) − 2 3 x + x x 2 − 4 ] ⋅ ( x + 2 x ) 2 3 ( x 2 − 4 ) \left\lbrack\frac{2}{3\left(x+2\right)}-\frac{2}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} [ 3 ( x + 2 ) 2 − 3 x 2 + x 2 − 4 x ] ⋅ ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

[ 1 3 ( x + 2 ) − 1 3 x + 1 2 ( x 2 − 4 ) ] ⋅ ( x + 2 x ) 2 3 ( x 2 − 4 ) \left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{1}{2\left(x^2-4\right)}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)} [ 3 ( x + 2 ) 1 − 3 x 1 + 2 ( x 2 − 4 ) 1 ] ⋅ ( x x + 2 ) 3 2 ( x 2 − 4 ) <path d="M983 90 l0 -0 c4,-6.7,10,-10,18,-10 H400000v40 H1013.1s-83.4,268,-264.1,840c-180.7,572,-277,876.3,-289,913c-4.7,4.7,-12.7,7,-24,7 s-12,0,-12,0c-1.3,-3.3,-3.7,-11.7,-7,-25c-35.3,-125.3,-106.7,-373.3,-214,-744 c-10,12,-21,25,-33,39s-32,39,-32,39c-6,-5.3,-15,-14,-27,-26s25,-30,25,-30 c26.7,-32.7,52,-63,76,-91s52,-60,52,-60s208,722,208,722 c56,-175.3,126.3,-397.3,211,-666c84.7,-268.7,153.8,-488.2,207.5,-658.5 c53.7,-170.3,84.5,-266.8,92.5,-289.5z M1001 80h400000v40h-400000z">

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Logarithmic Differentiation

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Logarithmic Differentiation

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Multiple Choice

Use logarithmic differentiation to find the derivative of the given function.
$y=\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

$\left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

$\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{x}{x^2-4}$

$\left\lbrack\frac{2}{3\left(x+2\right)}-\frac{2}{3x}+\frac{x}{x^2-4}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

$\left\lbrack\frac{1}{3\left(x+2\right)}-\frac{1}{3x}+\frac{1}{2\left(x^2-4\right)}\right\rbrack\cdot\sqrt{\left(\frac{x+2}{x}\right)^{\frac23}\left(x^2-4\right)}$

Verified step by step guidance

Step 1: Start by taking the natural logarithm of both sides of the equation to simplify the differentiation process. This gives us ln(y) = ln(\sqrt{((x+2)/x)^{2/3}(x^2-4)}).

Step 2: Use the properties of logarithms to simplify the right-hand side. The expression becomes ln(y) = (1/2) * ln(((x+2)/x)^{2/3}) + (1/2) * ln(x^2-4).

Step 3: Further simplify using the power rule for logarithms: ln(y) = (1/3) * ln((x+2)/x) + (1/2) * ln(x^2-4).

Step 4: Differentiate both sides with respect to x. The left side becomes (1/y) * dy/dx. For the right side, use the chain rule and the quotient rule where necessary.

Step 5: Solve for dy/dx by multiplying both sides by y. Substitute back the original expression for y to express the derivative in terms of x.

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