Find the derivative of the function.y=cos3(secθ)y=\cos^3\left(\sec\theta\right)y=cos3(secθ)

− 3 sec ⁡ θ tan ⁡ θ ⋅ sin ⁡ ( sec ⁡ θ ) cos ⁡ 2 ( sec ⁡ θ ) -3\sec\theta\tan\theta\cdot\sin\left(\sec\theta\right)\cos^2\left(\sec\theta\right)

Find the derivative of the function. y = cos 3 ( sec θ ) y=\cos^3\left(\sec\theta\right) y = cos 3 ( sec θ )

this problem, we're asked to find the derivative of the function y equals cosine cubed of the secant of theta. Now remember that whenever we're dealing with a trig function raised to a power, it's gonna be really helpful for us to go ahead and pull that power to the outside. So I'm gonna go ahead and do that here, rewriting my function as y equals cosine of secant theta, and all of this is raised to the power of 3. Now it's gonna be much more clear how exactly we're going to apply the chain rule. So let's go ahead and take a look at our functions, breaking them down to our outermost function and down to our innermost function. So our innermost function here is that secant of theta. Then as we go outward, we have a cosine. And then the furthest outward, we have all of this raised to the power of 3. So from here, we can go ahead and apply our chain rule to find our derivative y prime. Now to find our derivative here, we're starting from the outside, working our way in. Our outermost function is this cubed. So we're gonna pull that 3 out to the front using our power rule, decrease that exponent by 1. So I have 3, and then all of that stuff inside is staying the exact same as is, cosine of secant of theta. Nothing to worry about there. Cosine of secant of theta. And then 3 minus 1 is going to give us 2. Now from here, we're working our way in. We're moving on to our next function as we move inward, and that is going to be our cosine function. Now what is the derivative of a cosine? It is a negative sine. So we're multiplying here by the negative sine, and we're keeping that argument the same because remember whatever is inside of our function that we're working with is going to remain the same. So we have the negative sign of the secant of theta. Now moving to our furthest inward function, we are left with the secant of theta. What is the derivative of the secant of theta? Well, it is the tangent times the secant. So this is tangent theta times secant theta, and we have fully taken that derivative. Now we can do a little bit of simplification here by just pulling that 2 back to the inside and reordering a bit of this stuff. So I'm gonna just pull my, my trig functions that only have an argument of theta to the front just because they're the most simple functions. So I have 3 tangent theta secant theta, and then I have this negative sign of secant theta. I'm gonna go ahead and pull that negative out to the front to account for it. So times the sine of the secant of theta. And then finally, my last function I need to account for is the cosine squared of the secant theta. Make sure that you don't forget that last power. So this is my final answer for my derivative. Even if you wrote it in a different order, it is still an equivalent statement. This is just the order that by convention you would probably see this answer written in. So this is my final answer here. Thanks for watching and I'll see you in the next one.

Find the derivative of the function. y = cos ⁡ 3 ( sec ⁡ θ ) y=\cos^3\left(\sec\theta\right) y = cos 3 ( sec θ )

− 3 sec ⁡ θ tan ⁡ θ ⋅ sin ⁡ ( sec ⁡ θ ) cos ⁡ 2 ( sec ⁡ θ ) -3\sec\theta\tan\theta\cdot\sin\left(\sec\theta\right)\cos^2\left(\sec\theta\right)

3 cos ⁡ 2 ( sec ⁡ θ ) sec ⁡ θ tan ⁡ θ 3\cos^2\left(\sec\theta\right)\sec\theta\tan\theta 3 cos 2 ( sec θ ) sec θ tan θ

− 3 sin ⁡ 2 ( sec ⁡ θ tan ⁡ θ ) -3\sin^2\left(\sec\theta\tan\theta\right) − 3 sin 2 ( sec θ tan θ )

− 3 cos ⁡ 2 ( sec ⁡ θ tan ⁡ θ ) -3\cos^2\left(\sec\theta\tan\theta\right) − 3 cos 2 ( sec θ tan θ )

3. Techniques of Differentiation

The Chain Rule

Calculus

3. Techniques of Differentiation

The Chain Rule

Struggling with Calculus?

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Multiple Choice

Find the derivative of the function.
$y=\cos^3\left(\sec\theta\right)$

$3\cos^2\left(\sec\theta\right)\sec\theta\tan\theta$

$- 3 \sec θ \tan θ \cdot \sin (\sec θ) \cos^{2} (\sec θ)$

$-3\sin^2\left(\sec\theta\tan\theta\right)$

$-3\cos^2\left(\sec\theta\tan\theta\right)$

Verified step by step guidance

Identify the function y = \cos^3(\sec\theta). This is a composite function where the outer function is u^3 and the inner function is u = \cos(\sec\theta).

Apply the chain rule to find the derivative. The chain rule states that if y = f(g(x)), then the derivative y' = f'(g(x)) * g'(x).

Differentiate the outer function u^3 with respect to u, which gives 3u^2. Substitute u = \cos(\sec\theta) back into this derivative, resulting in 3\cos^2(\sec\theta).

Differentiate the inner function \cos(\sec\theta) with respect to \theta. Use the chain rule again: the derivative of \cos(x) is -\sin(x), and the derivative of \sec\theta is \sec\theta\tan\theta. Therefore, the derivative of \cos(\sec\theta) is -\sin(\sec\theta) * \sec\theta\tan\theta.

Combine the derivatives from the previous steps using the chain rule: y' = 3\cos^2(\sec\theta) * (-\sin(\sec\theta) * \sec\theta\tan\theta). Simplify the expression to get the final derivative.

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