Math  /  Calculus

Question9. Find dydθ\frac{d y}{d \theta} for y=cscθcotθy=\csc \theta-\cot \theta. (a) 0 (b) cot2θ+cscθcotθ-\cot ^{2} \theta+\csc \theta \cot \theta (c) secθtanθsec2θ\sec \theta \tan \theta-\sec ^{2} \theta (d) cscθcotθ+csc2θ-\csc \theta \cot \theta+\csc ^{2} \theta (e) None of these

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of y=cscθcotθy = \csc\theta - \cot\theta with respect to θ\theta. Watch out! Don't mix up the derivatives of cscθ\csc\theta and cotθ\cot\theta.
Remember those negative signs!

STEP 2

1. Rewrite in terms of sine and cosine
2. Compute the derivative
3. Simplify the result

STEP 3

Remember that cscθ\csc\theta is the reciprocal of sinθ\sin\theta, so we can rewrite it as 1sinθ\frac{1}{\sin\theta}.
This makes it easier to differentiate!

STEP 4

Similarly, cotθ\cot\theta is cosθsinθ\frac{\cos\theta}{\sin\theta}.
Now our function looks like this: y=1sinθcosθsinθ.y = \frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}.

STEP 5

Since both terms have the same denominator, we can combine them: y=1cosθsinθ.y = \frac{1-\cos\theta}{\sin\theta}.

STEP 6

Let's use the quotient rule, which says the derivative of f(θ)g(θ)\frac{f(\theta)}{g(\theta)} is g(θ)f(θ)f(θ)g(θ)(g(θ))2\frac{g(\theta) \cdot f'(\theta) - f(\theta) \cdot g'(\theta)}{(g(\theta))^2}.
Here, f(θ)=1cosθf(\theta) = 1 - \cos\theta and g(θ)=sinθg(\theta) = \sin\theta.

STEP 7

The derivative of 1cosθ1 - \cos\theta is sinθ\sin\theta, and the derivative of sinθ\sin\theta is cosθ\cos\theta.

STEP 8

dydθ=sinθsinθ(1cosθ)cosθ(sinθ)2=sin2θcosθ+cos2θsin2θ.\frac{dy}{d\theta} = \frac{\sin\theta \cdot \sin\theta - (1-\cos\theta) \cdot \cos\theta}{(\sin\theta)^2} = \frac{\sin^2\theta - \cos\theta + \cos^2\theta}{\sin^2\theta}.

STEP 9

Remember the super useful identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1!
We can substitute that into our derivative: dydθ=1cosθsin2θ.\frac{dy}{d\theta} = \frac{1 - \cos\theta}{\sin^2\theta}.

STEP 10

We can rewrite sin2θ\sin^2\theta as sinθsinθ\sin\theta \cdot \sin\theta.
Then, we have: dydθ=1cosθsinθsinθ=1cosθsinθ1sinθ.\frac{dy}{d\theta} = \frac{1 - \cos\theta}{\sin\theta \cdot \sin\theta} = \frac{1-\cos\theta}{\sin\theta} \cdot \frac{1}{\sin\theta}.

STEP 11

Notice that 1cosθsinθ\frac{1-\cos\theta}{\sin\theta} can't be simplified further with our given options.
However, 1sinθ\frac{1}{\sin\theta} is cscθ\csc\theta, so we get: dydθ=1cosθsinθcscθ.\frac{dy}{d\theta} = \frac{1-\cos\theta}{\sin\theta}\csc\theta. This isn't one of the options, so the answer is (e).

STEP 12

The derivative is 1cosθsinθcscθ\frac{1-\cos\theta}{\sin\theta}\csc\theta, which corresponds to answer **(e) None of these**.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord