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

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

Here we're asked to find the derivative of the function f of x equals four x 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 four x squared multiplied by the secant of x. And here, I'm subtracting the square root of x. I'm gonna 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 one half. So that's really what's being subtracted here. If I rewrite my function, four x secant x minus x to the power of one 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, four x squared secant x. Now focusing in on this term here and finding our derivative f prime of x, because I have two 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, four x squared, and we multiply that by the derivative of the right function, d d x 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 four x 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 one 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 four x 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 four x squared, which I, of course, want to go ahead and use the power rule here, multiplying by that two, decreasing that power by one. That's going to give me eight x, just eight x because two minus one is one. I don't need to worry about writing that. And then finally subtracting the derivative of that last term, x to the power of one half. Again, applying the power rule here, multiplying by that one half and then decreasing that power by one, that's going to give me one 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, four x squared tangent x secant x. 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 one half x to the power of negative one half. Because we started out with that square root, I'm gonna 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 one over two 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 four x squared, and I have this eight x. And both of those have four x in common, so that's something that I could go ahead and factor out. So if I factor out that four x, that's going to become x tangent x secant x plus two secant x having factored out just a four x from that term, and then still minus one over two 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} f ( x ) = 4 x 2 sec x − 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">

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}} 4 x 2 sec x tan 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 + 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">

12. Trigonometric Functions

Derivatives of Trig Functions

Business Calculus

12. Trigonometric Functions

Derivatives of Trig Functions

Struggling with Business Calculus?

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Multiple Choice

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

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

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

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

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

Verified step by step guidance

Step 1: Identify the function to differentiate. The given function is f(x) = 4x^2 sec(x) - sqrt(x). This function consists of two terms: 4x^2 sec(x) and -sqrt(x). We will differentiate each term separately using the rules of differentiation.

Step 2: Differentiate the first term, 4x^2 sec(x). Use the product rule since this term is a product of 4x^2 and sec(x). The product rule states that if u(x) and v(x) are functions, then d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). Here, u(x) = 4x^2 and v(x) = sec(x). Differentiate u(x) to get u'(x) = 8x, and differentiate v(x) to get v'(x) = sec(x)tan(x). Substitute these into the product rule formula.

Step 3: Differentiate the second term, -sqrt(x). Recall that sqrt(x) can be written as x^(1/2). Use the power rule for differentiation, which states that d/dx[x^n] = n*x^(n-1). Here, n = 1/2, so the derivative of -sqrt(x) is -1/2 * x^(-1/2). Simplify this to -1/(2sqrt(x)).

Step 4: Combine the results from Steps 2 and 3. The derivative of the first term, 4x^2 sec(x), is 8x sec(x) + 4x^2 sec(x)tan(x). The derivative of the second term, -sqrt(x), is -1/(2sqrt(x)). Add these together to get the derivative of the entire function.

Step 5: Write the final expression for the derivative. Combine all terms to get f'(x) = 4x^2 sec(x)tan(x) + 8x sec(x) - 1/(2sqrt(x)). This is the derivative of the given function.

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