Find the derivative of the given function.h(x)=ln(x+1x2+3)h\left(x\right)=\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)h(x)=ln(x2+3x+1)

−3x2−4x+32(x+1)(x2+3)\frac{$$-3x^2-4x+3$$}{2\left(x+1\right)\$$left(x^2+3$$\right)}2(x+1)(x2+3)−3x2−4x+3

Find the derivative of the given function. h ( x ) = ln ( x + 1 x 2 + 3 ) h\left(x\right)=\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right) h ( x ) = ln ( x 2 + 3 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"> )

In this problem, we wanna find the derivative of the function h of x equals the natural log of the square root of x plus 1 over x squared plus 3. Now in solving this problem, since my argument is not just x, it is actually a fraction of one function being divided by another function, I'm going to have to use here both the chain rule and the quotient rule. Now most of you should have learned those rules at this point. But if you haven't, you're gonna wanna circle back to this practice problem after you've already learned your chain and quotient rules. So let's go ahead and get into this for those of you that know these rules already. Here, we're finding our derivative here, h prime of x. Starting with our outermost function, that's what we wanna do using the chain rule. Our outermost function is the natural log. So the derivative of the natural log is a one over whatever that argument is. Now here, since I have a fraction, I can really just flip this fraction. So I'm just taking the reciprocal. That's what one over this is going to be. So this gives me x squared plus 3 over the square root of x plus 1. Now I'm then multiplying by the derivative of that inside function. In this case, since I have a fraction, I have one function being divided by another function. I'm going to apply the quotient rule as I'm applying this chain rule. So we want to multiply this by the derivative of the square root of x+1 over x squared plus 3. Remember our quotient rule memory tool, low d high minus high d low over the square of what's below. So that low function, that's x squared plus 3 times the derivative of that high function. It's going to be helpful here to think of this as x plus 1 to the power of 1 half. So then this derivative is a 1 half times x plus 1 to the power of negative 1 half minus high d low, so that's the square root of x plus 1, times the derivative of x squared plus 3, which is just 2x over the square of what's below. So that's x squared plus 3 squared. So here, I've ex I have successfully applied both the chain rule and the quotient rule while I was doing that chain rule. So that means all of our calculus is actually done here. So it remains to just do a lot of algebra, which can often be the trickiest part because it's really easy to just miss a term here or there. So make sure that you're counting for every single term every time you go to a new line. So let's go ahead and figure out exactly what's going on here. For now, I'm going to leave this term alone and focus mostly on this numerator here. Now looking at this term, this x squared plus 3 times 1 half times x plus 1 to the power of negative one half, I can rewrite this as x squared plus 3 over 2 root x plus 1. So I'm going to keep everything else the same here. That's x squared plus 3 over root x plus 1, making sure I'm accounting for everything here, minus 2x times root x plus 1 over x squared plus 3 squared. Now looking at this, since I have a fraction and then not a fraction right here, I'm going to go ahead and give these common denominators so that I can actually subtract them. So since this denominator is 2 times rootx plus 1, I'm going to multiply by 2 root x plus 1 over 2 root x plus 1. Since here, I'm effectively just multiplying by 1, I'm not actually changing the value of everything. I'm just making it easier for me to simplify from here. So if I go ahead and rewrite this, this is x squared plus 3 and then multiplying multiplying 2x times root x plus 1 times 2 root x plus 1. This is 4x times x plus 1 because since I have 2 square roots being multiplied the same thing underneath the square roots, that square root goes away because it's really just being squared. All of this is over 2 root x plus 1. That's that common denominator that we found. Then this is over again, x squared plus 3 squared, and keeping this other term the same, x squared plus 3 over rootx plus 1. So let's organize what exactly is going on here. Now I have a fraction multiplying a fraction within a fraction. Because I have a fraction within a fraction, I really just want to move this denominator down to be with my existing denominator. That is totally fine to do here. So I'm going to still keep this term the same, x squared plus 3 over root x plus 1. I'm just rewriting it so that I don't forget about it later. Then in the top, I'm going to go ahead and distribute this 4x into my parentheses. That gives me x squared plus 3 minus 4x squared minus 4x divided by everything now that is in my denominator. That's 2 times rootx+one timesxsquaredplus3squared. So that's everything. That's 2 rootx+1 and then x squared plus 3 squared. So all of that is now in my existing denominator. Now from here, we can combine some like terms. I have x squared, and I have minus 4x squared. Then, also, I have this x squared plus 3, and I have x squared plus 3 squared on the bottom, so I can get one of those to cancel out. Then also on the bottom, I have root x plus 1 multiplying another root x plus 1. That means I'm just squaring that term. So let's rewrite this with all of those changes. So x squared minus 4x squared, that gives me negative 3x squared. Writing this in standard form, minus 4x plus 3. So that's just ascending order of exponents. Then remember, we still have this 2. And then we have a root x plus 1. And that's being squared because I have 2 of them here. And then finally, I have x squared plus 3. That is no longer squared because I canceled one of them out. Now my very last step here, I can go ahead and cancel that square with that square root, leaving me with my final answer as negative 3x squared minus 4x plus 3 over 2 times x plus 1 with that square root taken away times x squared plus 3. So this is my final answer here. This is my final derivative. Remember to make sure and stay organized with all of those terms floating around and stay clear on that algebra. Let us know if you have any questions here, and let's continue on.

Find the derivative of the given function.h(x)=ln⁡(x+1x2+3)h\left(x\right)=\ln\left(\frac{\sqrt{x+1}}{$$x^2+3$$}\right)h(x)=ln(x2+3x+1)

−3x2−4x+32(x+1)(x2+3)\frac{$$-3x^2-4x+3$$}{2\left(x+1\right)\$$left(x^2+3$$\right)}2(x+1)(x2+3)−3x2−4x+3

x2+3x+1\frac{$$x^2+3$$}{\sqrt{x+1}}x+1x2+3

−3x2−4x+32xx+1⋅(x2+3)2\frac{$$-3x^2-4x+3$$}{2x\sqrt{x+1}\cdot\$$left(x^2+3$$\$$right)^2$$}2xx+1⋅(x2+3)2−3x2−4x+3

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Business Calculus

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

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Multiple Choice

Find the derivative of the given function.
$h$\left$(x$\right$)=$\ln\]\left$($\frac{\sqrt{x+1}$}{x^2+3}$\right$)$

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

$$\frac{x^2+3}{\sqrt{x+1}$}$

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

$$\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}\]\cdot\[\ln\]\left$($\frac{\sqrt{x+1}$}{x^2+3}$\right$)$

Verified step by step guidance

Step 1: Recognize that the function h(x) = ln((sqrt(x+1))/(x^2+3)) is a logarithmic function. To differentiate it, use the chain rule and the properties of logarithms. Start by simplifying the logarithmic expression using the property ln(a/b) = ln(a) - ln(b). This gives h(x) = ln(sqrt(x+1)) - ln(x^2+3).

Step 2: Further simplify ln(sqrt(x+1)) using the property ln(a^b) = b*ln(a). Since sqrt(x+1) is equivalent to (x+1)^(1/2), we can rewrite ln(sqrt(x+1)) as (1/2)*ln(x+1). Thus, h(x) = (1/2)*ln(x+1) - ln(x^2+3).

Step 3: Differentiate each term separately. For the first term (1/2)*ln(x+1), use the derivative of ln(u), which is 1/u * du/dx. Here, u = x+1, so the derivative is (1/2)*(1/(x+1)). For the second term ln(x^2+3), use the same rule, where u = x^2+3. The derivative is (1/(x^2+3)) * d(x^2+3)/dx.

Step 4: Compute the derivative of x^2+3. The derivative of x^2 is 2x, and the derivative of 3 is 0. Therefore, d(x^2+3)/dx = 2x. Substitute this back into the derivative of ln(x^2+3), which becomes (1/(x^2+3)) * 2x.

Step 5: Combine the results from Step 3 and Step 4. The derivative of h(x) is h'(x) = (1/2)*(1/(x+1)) - (2x/(x^2+3)). Simplify further if needed to express the derivative in its final form.

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Struggling with Business Calculus?

Find the derivative of the given function.h(x)=ln(x+1x2+3)h\(\left\)(x\(\right\))=\(\ln\]\left\)(\(\frac{\sqrt{x+1}\)}{x^2+3}\(\right\))h(x)=ln(x2+3x+1)

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Derivatives of Exponential & Logarithmic Functions practice set

Find the derivative of the given function.
$h\(\left\)(x\(\right\))=\(\ln\]\left\)(\(\frac{\sqrt{x+1}\)}{x^2+3}\(\right\))$