Use logarithmic differentiation to find the derivative of the given function. y = ( x + 1 ) x y=(\sqrt{x+1})^{x} y = ( x + 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> ) x

Here, we wanna use log differentiation to find the derivative of the given function. Here, we're given y equals the square root of x plus 1 all raised to the power of x. Now in order to use log differentiation, it can be a good idea to rewrite anything that can be a power as a power. Whenever we're taking the square root of something, that's the same thing as raising it to the power of 1 half. So I can rewrite this as x plus 1 to the power of x over 2, just combining that 1 half with that existing power of x. So from here, let's go ahead and get started with our log differentiation by taking the natural log on each side. So this is the natural log of y is equal to the natural log of x plus 1 to the power of x over 2. Now since we're taking the natural log of all of this stuff, we can pull that power out front and multiply by it. So that then gives me x over 2 times the natural log of x plus 1. And this is equal to just the natural log of y. Now from here, we can proceed in using implicit differentiation. That means we're just taking a ddx, our derivative, on each side here, making sure we know we're taking the derivative of all of that. On this left side, ddx of the natural log of y, this is typically how these log differentiation problems work out. This is one over y times dydx, having applied the chain rule there. Then over here on my right hand side, I have to use the product rule because I have x over 2 multiplying the natural log of x plus 1. So remember our rule for or our memory tool for remembering the product rule, left d right plus a right d left. So I take that left hand function x over 2, multiply it by the derivative of my right hand function. The derivative of the natural log of x plus 1 is 1 over x plus 1. Now I do need to think about using the chain rule, but the derivative of x plus 1 is just 1, so it won't change anything here. Continuing on with our product rule here, adding right d left, so that's the natural log of x plus 1, leaving that right hand function as is, multiplying it by the derivative of x over 2. This derivative is just a one half. So from here, we want dy dx by itself. We've taken all the derivatives we need to. We just need to get that dy dx isolated and get this all in terms of x. Now, of course, we wanna multiply both sides by y to get that to cancel, leaving me with just dydx. That's what I want to find here. And then this becomes x over 2. I'm gonna go ahead and combine all these fractions into 1. So this is x over 2, x plus 1, just condensing that a little bit, plus the natural log of x+2. And all of this is getting multiplied by y, and we know what y is. It is x+one to the power of x over 2, and this is my final derivative here.

Use logarithmic differentiation to find the derivative of the given function. y = ( x + 1 ) x y=(\sqrt{x+1})^{x} y = ( x + 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ) x

[ x 2 ( x + 1 ) + ln ⁡ ( x + 1 ) 2 ] ( x + 1 ) x 2 \left\lbrack\frac{x}{2\left(x+1\right)}+\frac{\ln\left(x+1\right)}{2}\right\rbrack\left(x+1\right)^{\frac{x}{2}}

x ( x + 1 ) x − 1 x\left(\sqrt{x+1}\right)^{x-1} x ( x + 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ) x − 1

x 2 ( x + 1 ) x 2 − 1 \frac{x}{2}\left(\sqrt{x+1}\right)^{\frac{x}{2}-1} 2 x ( x + 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ) 2 x − 1

[ ln ⁡ ( x + 1 ) + x 2 ( x + 1 ) ] ( x + 1 ) x \left\lbrack\frac{\ln\left(x+1\right)+x}{2\left(x+1\right)}\right\rbrack\left(\sqrt{x+1}\right)^{x} [ 2 ( x + 1 ) l n ( x + 1 ) + x ] ( x + 1 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ) x

5. Applications of Derivatives

Logarithmic Differentiation

Business Calculus

5. Applications of Derivatives

Logarithmic Differentiation

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Multiple Choice

Use logarithmic differentiation to find the derivative of the given function.
$y=(\sqrt{x+1})^{x}$

$x\left(\sqrt{x+1}\right)^{x-1}$

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

$[\frac{x}{2 (x + 1)} + \frac{\ln (x + 1)}{2}] {(x + 1)}^{\frac{x}{2}}$

$\left\lbrack\frac{\ln\left(x+1\right)+x}{2\left(x+1\right)}\right\rbrack\left(\sqrt{x+1}\right)^{x}$

Verified step by step guidance

Step 1: Begin by identifying the function y = (sqrt(x+1))^(x/2). Since the function involves both a variable base and a variable exponent, logarithmic differentiation is a useful method to simplify the differentiation process.

Step 2: Take the natural logarithm of both sides of the equation to simplify the expression. This gives ln(y) = ln((sqrt(x+1))^(x/2)). Using logarithmic properties, rewrite this as ln(y) = (x/2) * ln(sqrt(x+1)).

Step 3: Simplify further using the property ln(sqrt(x+1)) = (1/2) * ln(x+1). Substituting this back, we get ln(y) = (x/2) * (1/2) * ln(x+1), which simplifies to ln(y) = (x/4) * ln(x+1).

Step 4: Differentiate both sides of the equation with respect to x. On the left-hand side, the derivative of ln(y) is (1/y) * dy/dx. On the right-hand side, use the product rule to differentiate (x/4) * ln(x+1).

Step 5: After applying the product rule, solve for dy/dx by multiplying through by y. Substitute y = (sqrt(x+1))^(x/2) back into the equation to express the derivative in terms of the original function.