Find the derivative of the function.f(x)=4x2secx−xf\left(x\right)=4x^2\sec x-\sqrt{x}

4 x 2 sec ⁡ x tan ⁡ x + 8 x sec ⁡ x − 1 2 x 4x^2\sec x\tan x+8x\sec x-\frac{1}{2\sqrt{x}}

Find the derivative of the function. f ( x ) = 4 x 2 sec x − x f\left(x\right)=4x^2\sec x-\sqrt{x}

we're asked to find the derivative of the function f of x equals 4x squared secant x minus the square root of x. Now here, we have a couple of different things going on. We have some functions being multiplied. I have this 4x squared multiplied by the secant of x. And here, I'm subtracting the square root of x. I'm going to go ahead and rewrite my square root of x as a power function because I know that will make it easier to use my power rule. We know that the square root of x is x to the power of 1 half. So that's really what's being subtracted here if I rewrite my function, 4xsecantx minus x to the power of 1 half. Now, in finding the derivative here, we're going to have to use a couple of different rules. I'm going to go ahead and break down for you exactly what rules that I'm using. But just know as you continue to take more and more derivatives, you're not going to need to break it down like that. If you can go directly from your original function straight to your derivative answer, that's totally fine. Here, I'm just gonna break it down for you so you can see exactly what's happening mathematically. So looking at this first term, 4x squared secant x. Now focusing in on this term here and finding our derivative, f prime of x, because I have 2 functions that are being multiplied together, I need to use the product rule. Now the product rule tells us that if we take that left side function, 4x squared, and we multiply that by the derivative of the right function, ddx of secant squared, and we add that together with that right side function, which here again is secant of x, and then multiply that by the derivative ddx of 4x squared. That's going to give us our derivative. But remember that we right now are focused in on that first term. We still need to take the derivative of that second term as well. Now here, I'm going to add this or I'm going to subtract because I have subtraction happening here using my difference rule for derivatives. I'm subtracting ddx of that x to the power of 1 half. So here, you can see all of the different rules that I'm using. These two first terms come from using the product rule. And then here, I apply the difference rule as well in order to get that final derivative. Now here, we're actually going to take all of these derivatives, these d d x's that we see in here, using our different rules. So I have that 4x squared that's staying the same. The derivative of the secant of x is going to be the tangent of x times the secant of x. And then I'm adding that together with the secant of x times the derivative of 4x squared, which I, of course, want to go ahead and use the power rule here, multiplying by that 2, decreasing that power by 1. That's going to give me 8x just 8x because 2 minus 1 is 1. I don't need to worry about writing that. And then finally, subtracting the derivative of that last term, x to the power of 1 half, again, applying the power rule here, multiplying by that 1 half and then decreasing that power by 1, that's going to give me 1 half x to the power of negative one half. Now, this is an acceptable answer here. But we, of course, want to do a little bit of simplification and write this in a little bit better, more organized way. So looking at that first term, I'm going to keep that as is, 4x squared tangentx secantx. And then I'm adding that together. I'm going to go ahead and move that 8x to the front. Typically, when you are writing trig functions, you're going to want to put constants and other variables in front of it. So I have 8x secant x. Then I'm subtracting here 1 half x to the power of negative one half. Because we started out with that square root, I'm going to change this back to a square root. But remember that because it is a negative power, it's going to go on the bottom. So this is minus 1 over 2 root x. Now, this again is an acceptable answer. This is correct. This is a correct derivative. But remember that your professor or your instructor might want you to simplify more. So if you are asked to simplify as much as possible, there are a couple of things that we can do here just based on some common terms. I have this 4x squared, and I have this 8x. And both of those have 4x in common, so that's something that I could go ahead and factor out. So if I factor out that 4x, that's going to become x tangentxsecantxplus2secantxhavingfactored out just a 4x from that term, and then still minus 1 over 2 root x. Now this is also something to keep in mind if you're dealing with a multiple choice question. Even though you might not see the answer that you got, remember that they might have simplified differently. So always double check that there's not something more that you could do here in order to make sure that you get the right answer every single time. So both of these answers are acceptable. This is our final derivative of our function and feel free to let me know if you have questions here.

Find the derivative of the function. f ( x ) = 4 x 2 sec ⁡ x − x f\left(x\right)=4x^2\sec x-\sqrt{x}

4 x 2 sec ⁡ x tan ⁡ x + 8 x sec ⁡ x − 1 2 x 4x^2\sec x\tan x+8x\sec x-\frac{1}{2\sqrt{x}}

8 x sec ⁡ x + 1 2 x 8x\sec x+\frac{1}{2\sqrt{x}} 8 x sec x + 2 x <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"> 1

8 x sec ⁡ x tan ⁡ x − 1 2 x 8x\sec x\tan x-\frac{1}{2\sqrt{x}} 8 x sec x tan x − 2 x <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"> 1

4 x 2 sec ⁡ x tan ⁡ x + 8 x sec ⁡ x − 1 2 x 4x^2\sec x\tan x+8x\sec x-\frac12\sqrt{x} 4 x 2 sec x tan x + 8 x sec x − 2 1 x <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">

3. Techniques of Differentiation

Derivatives of Trig Functions

Calculus

3. Techniques of Differentiation

Derivatives of Trig Functions

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Multiple Choice

Find the derivative of the function.
$f (x) = 4 x^{2} \sec x - \sqrt{x}$

$8x\sec x+\frac{1}{2\sqrt{x}}$

$8x\sec x\tan x-\frac{1}{2\sqrt{x}}$

$4x^2\sec x\tan x+8x\sec x-\frac12\sqrt{x}$

$4 x^{2} \sec x \tan x + 8 x \sec x - \frac{1}{2 \sqrt{x}}$

Verified step by step guidance

Identify the function to differentiate: f(x) = 4x^2 sec(x) - sqrt(x).

Apply the product rule to the term 4x^2 sec(x). The product rule states that if you have a function u(x)v(x), its derivative is u'(x)v(x) + u(x)v'(x). Here, let u(x) = 4x^2 and v(x) = sec(x).

Differentiate u(x) = 4x^2 to get u'(x) = 8x.

Differentiate v(x) = sec(x) to get v'(x) = sec(x)tan(x).

Apply the power rule to differentiate sqrt(x), which is x^(1/2). The derivative is (1/2)x^(-1/2).

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