Find the derivative of the function.f(x)=5cosx2x3f\left(x\right)=\frac{5\cos x}{2x^3}f(x)=2x35cosx

− 5 ( x sin ⁡ x + 3 cos ⁡ x ) 2 x 4 \frac{-5\left(x\sin x+3\cos x\right)}{2x^4} 2 x 4 − 5 ( x s i n x + 3 c o s x )

Find the derivative of the function. f ( x ) = 5 cos x 2 x 3 f\left(x\right)=\frac{5\cos x}{2x^3} f ( x ) = 2 x 3 5 c o s x

this problem, we're asked to find the derivative of the function f of x equals five cosine x over two x cubed. So I have a trig function present here, and I also am dealing with a function divided by a function, meaning that I need to use the quotient rule. So let's go ahead and get started in finding this derivative using that quotient rule. So we're finding f prime of x here using our quotient rule. Remember that our quotient rule tells us low d high minus high d low over the square of what's below. Remember that that rhymes. That will help you remember what exactly you should be doing on the top there. So we're doing a low d high. So I'm taking my low function, two x cubed, that's my function on the bottom, times the derivative of that top function. Now the derivative of five cosine x, I know that the derivative of cosine x is a negative sine, and that constant will stay as is. So this will be negative five sine of x. Then I am subtracting here high d low. So that high function, my function on top, five cosine x times the derivative of that bottom function, two x cubed, using the power rule, this will give me 6x squared. Then I'm dividing all of this by the square of that bottom function, the square of what's below. So I have 2x cubed squared. Now, we've applied our quotient rule in full here, and we know that we just have a lot of simplification ahead of us. So let's go ahead and start simplifying here and expanding stuff out so that we can end up simplifying in the end. Now this first term here, two x cubed minus five times negative five sine x, we're gonna fully multiply that out. That will give us negative 10 x cubed sine x. And then that second term, multiplying that six x squared times that five cosine x, that will give us a minus 30 x squared cosine x. Then in the bottom, if we actually if we actually square this function to x cubed, remember that I need to square both the two and the x cubed. Squaring that two gives me four. And then if I square x cube, that's going to give me x to the power of six. Remember that when we raise a power to a power, we're multiplying them and not just adding them together. So we have simplified this out a little bit. And looking at this, I have some terms that are in common. Now in my top, I have x cubed and I have this x squared, and on the bottom, I have x to the power of six. So I can use some of those x's on the bottom in order to cancel what I have on the top. So all of these terms have at least an x squared in common. So I'm going to go ahead and divide that x squared out and cancel it. So this x squared will cancel. This x cubed will turn to just an x. And then this x to the power of six on the bottom will become x to the power of four because we have used two of those to cancel out those x squared. Now let's rewrite our function again here. We have a negative 10 x sine x minus 30 cosine x because we have fully canceled that x squared there. Then I'm dividing by four x to the power of four. Now depending on your instructor, they may want you to simplify differently. So make sure that you always double check with how your instructor specifically wants you to simplify when you get answers like this one. Now I can do a little bit more simplification here because I have all of these constants. I have negative 10, I have negative 30, and I have four. Now these all can be divided by two, so I can go ahead and cancel a two out of all of these. So this four will become two. This negative 10 will become a negative five. This negative 30 will become a negative 15. So this simplifies further to negative five x sine x minus 15 cosine x, all divided by two x to the power of four. Now we could stop here, but we can do one final thing because I have some common terms here. I have negative five and negative 15, so I can factor out a negative five from both of these terms, making this a negative five times x sine x, pulling out that negative five, plus three because I factored out negative five, plus three times cosine x, all divided by two x to the power of four. And this is as simplified as we can get here. We have canceled out all of our like terms, and this is our final fully simplified answer for the derivative of our original function. So we get negative five times x sine x plus three cosine x all over two x to the power of four. So remember, with a lot of our quotient rule problems, the hardest part is going to be fully simplifying our answer. And sometimes this process can be a bit tedious. Make sure that you're fully paying attention to all of your terms and carrying over what you need to when you're canceling out stuff. Don't forget any of those terms in order to get that final answer correct. Feel free to let me know if you have questions here, and I'll see you in the next one.

Find the derivative of the function. f ( x ) = 5 cos ⁡ x 2 x 3 f\left(x\right)=\frac{5\cos x}{2x^3} f ( x ) = 2 x 3 5 c o s x

− 5 ( x sin ⁡ x + 3 cos ⁡ x ) 2 x 4 \frac{-5\left(x\sin x+3\cos x\right)}{2x^4} 2 x 4 − 5 ( x s i n x + 3 c o s x )

− 5 ( x sin ⁡ x + 3 cos ⁡ x ) 2 x 3 \frac{-5\left(x\sin x+3\cos x\right)}{2x^3} 2 x 3 − 5 ( x s i n x + 3 c o s x )

5 x 2 cos ⁡ x − 10 x 3 sin ⁡ x 4 x 6 \frac{5x^2\cos x-10x^3\sin x}{4x^6} 4 x 6 5 x 2 c o s x − 10 x 3 s i n x

− 5 sin ⁡ x 6 x 2 \frac{-5\sin x}{6x^2} 6 x 2 − 5 s i n x

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.
$f\left(x\right)=\frac{5\cos x}{2x^3}$

$\frac{-5\left(x\sin x+3\cos x\right)}{2x^3}$

$\frac{-5\left(x\sin x+3\cos x\right)}{2x^4}$

$\frac{5x^2\cos x-10x^3\sin x}{4x^6}$

$\frac{-5\sin x}{6x^2}$

Verified step by step guidance

Step 1: Recognize that the given function is a quotient, so we will use the Quotient Rule to find the derivative. The Quotient Rule states: $ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} $, where $ u(x) $ is the numerator and $ v(x) $ is the denominator.

Step 2: Identify $ u(x) $ and $ v(x) $ from the function $ f(x) = \frac{5 \cos x}{2x^3} $. Here, $ u(x) = 5 \cos x $ and $ v(x) = 2x^3 $.

Step 3: Compute $ u'(x) $, the derivative of $ u(x) = 5 \cos x $. Using the derivative rule for cosine, $ \frac{d}{dx}(\cos x) = -\sin x $, we get $ u'(x) = -5 \sin x $.

Step 4: Compute $ v'(x) $, the derivative of $ v(x) = 2x^3 $. Using the power rule, $ \frac{d}{dx}(x^n) = n x^{n-1} $, we get $ v'(x) = 6x^2 $.

Step 5: Substitute $ u(x), u'(x), v(x), $ and $ v'(x) $ into the Quotient Rule formula. This gives: $ f'(x) = \frac{(-5 \sin x)(2x^3) - (5 \cos x)(6x^2)}{(2x^3)^2} $. Simplify the numerator and denominator to express the derivative in its simplest form.

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