Find the derivative of the function.y=cotθ3+secθy=\frac{\cot\theta}{3+\sec\theta}

− csc ⁡ 2 θ ( 3 + sec ⁡ θ ) − sec ⁡ θ ( 3 + sec ⁡ θ ) 2 \frac{-\csc^2\theta\left(3+\sec\theta\right)-\sec\theta}{\left(3+\sec\theta\right)^2}

Find the derivative of the function. y = cot θ 3 + sec θ y=\frac{\cot\theta}{3+\sec\theta}

In this problem, we're asked to find the derivative of the function y equals cotangent theta over 3 plus the secant of theta. Now here we have some trig functions. So we're going to apply those rules for taking derivatives of trig functions that we've just learned. And we also have division happening between these functions, which means that we need to use the quotient rule. Now, the quotient rule tells us a low d high minus high d low over the square of what's below. So that's exactly what we're going to use here to find this derivative, y prime. So starting with that first term, low d high. So I'm taking that bottom function, my low function, 3 plus secant theta. D high, that's the derivative of that top function, cotangent of theta, the derivative of the cotangent. This is a cofunction. We know absolutely it's going to be negative. And here it is a negative cosecant squared of theta. So that is low dhigh minushigh, so my cotangent of theta, d low, that's the derivative of that bottom function, 3+secanttheta. The derivative of 3+secanttheta, the derivative of that 3 doesn't matter. It's 0. So we really just need to worry about the derivative of secant theta. Now the derivative of secant of theta is tangent theta secant theta, and that's all that goes in that numerator. Now in my denominator over the square of what's below, 3+secant theta squared. Don't forget that square. That's a really easy thing to forget in these problems. And this is my derivative. This is a correct answer. But remember that based on how you simplify, your answer might be different than what you see in maybe a multiple choice answer or maybe what your instructor presents in an answer key maybe. Now that's okay because remember that you might have an equivalent expression, which would still be correct. Now if you're ever unsure, just go ahead and look fully at your expression to try to find any extra ways that you could simplify here. So looking at this, I do notice that I have a cotangent and a tangent. Now, what is cotangent? Well, remember that cotangent is one over the tangent of theta. So if I have one over the tangent of theta, and that's being multiplied by tangent of theta, those are just going to cancel out to 1. So here, this cotangent and tangent are just going to cancel because this is 1 over tangent times tangent. So using this simplification here, I can go ahead and rewrite my function. I'm gonna go ahead and move that negative cosecant squared to the front just to organize this a little bit better. This is negative cosecant squared of theta times 3 plus secant theta minus since I have canceled out cotangent and tangent, I am just left with that secant of theta. And all of this is getting divided by 3 plus secant theta squared. Now there may be some more and different ways to simplify this and that's okay because remember that your expression could be equivalent, but always, always double check. This is our correct answer here. The derivative of our function is negative cosecant squared of theta times 3 plus secant theta minus secanttheta all over 3+secanttheta2. Lots of secants in there. Now feel free to let us know if you have any questions, and I'll see you in the next one.

Find the derivative of the function. y = cot ⁡ θ 3 + sec ⁡ θ y=\frac{\cot\theta}{3+\sec\theta}

− csc ⁡ 2 θ ( 3 + sec ⁡ θ ) − sec ⁡ θ ( 3 + sec ⁡ θ ) 2 \frac{-\csc^2\theta\left(3+\sec\theta\right)-\sec\theta}{\left(3+\sec\theta\right)^2}

sec ⁡ θ + csc ⁡ 2 θ ( 3 + sec ⁡ θ ) ( 3 + sec ⁡ θ ) 2 \frac{\sec\theta+\csc^2\theta\left(3+\sec\theta\right)}{\left(3+\sec\theta\right)^2}

1 ( 3 + sec ⁡ θ ) 2 \frac{1}{\left(3+\sec\theta\right)^2}

− csc ⁡ 2 θ sec ⁡ θ tan ⁡ θ \frac{-\csc^2\theta}{\sec\theta\tan\theta}

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.
$y = \frac{\cot θ}{3 + \sec θ}$

$\frac{- \csc^{2} θ (3 + \sec θ) - \sec θ}{{(3 + \sec θ)}^{2}}$

$\frac{\sec θ + \csc^{2} θ (3 + \sec θ)}{{(3 + \sec θ)}^{2}}$

$\frac{1}{{(3 + \sec θ)}^{2}}$

$\frac{- \csc^{2} θ}{\sec θ \tan θ}$

Verified step by step guidance

Identify the function y = $\frac{\cot\theta}{3+\sec\theta}$. This is a quotient of two functions, so we will use the quotient rule to find the derivative.

Recall the quotient rule: if y = $\frac{u}{v}$, then y' = $\frac{u'v - uv'}{v^2}$. Here, u = $\cot\[\theta$ and v = 3 + $\sec\]\theta$.

Find the derivative of the numerator u = $\cot\[\theta$. The derivative of $\cot\]\theta$ is -$\csc$^2$\theta$.

Find the derivative of the denominator v = 3 + $\sec\[\theta$. The derivative of $\sec\]\theta$ is $\sec\[\theta\]\tan\[\theta$, so v' = $\sec\]\theta\[\tan\]\theta$.

Apply the quotient rule: y' = $\frac{(-\csc^2\theta)(3+\sec\theta) - (\cot\theta)(\sec\theta\tan\theta)}{(3+\sec\theta)^2}$. Simplify the expression to get the final derivative.

3:53m

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Struggling with Calculus?

Find the derivative of the function.y=cotθ3+secθy=\(\frac{\cot\theta}{3+\sec\theta}\)

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Derivatives of Trig Functions practice set

Find the derivative of the function.
$y = \frac{\cot θ}{3 + \sec θ}$