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

−5(xsin⁡x+3cos⁡x)2x4\frac{-5\left(x\sin x+3\cos x\right)}{$$2x^4$$}2x4−5(xsinx+3cosx)

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

this problem, we're asked to find the derivative of the function f of x equals 5 cosinex over 2x 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, 2x3, that's my function on the bottom, times the derivative of that top function. Now the derivative of 5 cosine x, I know that the derivative of cosx is a negative sine, and that constant will stay as is. So this will be negative 5 sine of x. Then I am subtracting here, high d low. So that high function, my function on top, 5 cosinex times the derivative of that bottom function, 2x 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, 2x cubed minus 5 times negative 5 sine x, we're gonna fully multiply that out. That will give us negative 10x cubed sine x. And then that second term, multiplying that 6x squared times that 5 cosine x, that will give us a minus 30 x squared cosine x. Then in the bottom, if we actually or if we actually square this function to x cubed, remember that I need to square both the 2 and the x cube. Squaring that 2 gives me 4. And then if I square x cube, that's going to give me x to the power of 6. 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 6. 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 6 on the bottom will become x to the power of 4 because we have used 2 of those to cancel out those x squared. Now let's rewrite our function again here. We have a negative 10xsinx minus 30 cosinex because we have fully canceled that x squared there. Then I'm dividing by 4x to the power of 4. 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 4. Now these all can be divided by 2, so I can go ahead and cancel a 2 out of all of these. So this 4 will become 2. This negative 10 will become a negative 5. This negative 30 will become negative 15. So this simplifies further to negative 5x sinex minus 15 cosinex, all divided by 2x to the power of 4. Now we could stop here, but we can do one final thing because I have some common terms here. I have negative 5 and negative 15. So I can factor out a negative 5 from both of these terms, making this a negative 5 times x sine x, pulling out that negative 5, plus 3, because I factored out negative 5, plus 3 times cosine x, all divided by 2x to the power of 4. 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 5 times x sine x plus 3 cosine x all over 2x to the power of 4. 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)=5cos⁡x2x3f\left(x\right)=\frac{5\cos x}{$$2x^3$$}

−5(xsin⁡x+3cos⁡x)2x4\frac{-5\left(x\sin x+3\cos x\right)}{$$2x^4$$}2x4−5(xsinx+3cosx)

−5(xsin⁡x+3cos⁡x)2x3\frac{-5\left(x\sin x+3\cos x\right)}{$$2x^3$$}2x3−5(xsinx+3cosx)

5x2cos⁡x−10x3sin⁡x4x6\frac{$$5x^2$$\cos $$x-10x^3$$\sin x}{$$4x^6$$}4x65x2cosx−10x3sinx

−5sin⁡x6x2\frac{-5\sin x}{$$6x^2$$}

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

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

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

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

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

Verified step by step guidance

Identify the function to differentiate: f(x) = $\frac{5\cos x}{2x^3}$ - $\frac{5(x\sin x + 3\cos x)}{2x^3}$.

Apply the quotient rule to differentiate each term separately. The quotient rule states that if you have a function $\frac{u}{v}$, its derivative is $\frac{u'v - uv'}{v^2}$.

For the first term $\frac{5\cos x}{2x^3}$, let u = 5$\cos$ x and v = 2x^3. Differentiate u and v separately: u' = -5$\sin$ x and v' = 6x^2.

For the second term $\frac{5(x\sin x + 3\cos x)}{2x^3}$, let u = 5(x$\sin$ x + 3$\cos$ x) and v = 2x^3. Differentiate u and v separately: u' = 5($\sin$ x + x$\cos$ x - 3$\sin$ x) and v' = 6x^2.

Combine the derivatives using the quotient rule for each term and simplify the expression. Remember to combine like terms and factor where possible.

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Struggling with Calculus?

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

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Derivatives of Trig Functions practice set

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