Get insightful responses to your questions quickly and easily on IDNLearn.com. Our platform provides detailed and accurate responses from experts, helping you navigate any topic with confidence.
Sagot :
To find the derivative [tex]\(\frac{dy}{dx}\)[/tex] of the function [tex]\( y = \frac{x^5 \cos^3 x}{\sqrt{\csc x}} \)[/tex], we will go through the steps systematically using the rules of differentiation. Here's the detailed step-by-step solution:
1. Rewrite the function in a more convenient form:
Given:
[tex]\[ y = \frac{x^5 \cos^3 x}{\sqrt{\csc x}} \][/tex]
Recall that [tex]\(\csc x = \frac{1}{\sin x}\)[/tex], so [tex]\(\sqrt{\csc x} = \sqrt{\frac{1}{\sin x}} = \frac{1}{\sqrt{\sin x}}\)[/tex].
Therefore, the function can be rewritten as:
[tex]\[ y = x^5 \cos^3 x \cdot \frac{1}{\sqrt{\sin x}} = \frac{x^5 \cos^3 x}{\sqrt{\sin x}} \][/tex]
2. Use the product rule and the chain rule:
To differentiate this expression, we will utilize the product rule:
[tex]\[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \][/tex]
where [tex]\(u = x^5 \cos^3 x\)[/tex] and [tex]\(v = \frac{1}{\sqrt{\sin x}}\)[/tex].
Let’s differentiate each part separately:
Differentiate [tex]\(u = x^5 \cos^3 x\)[/tex]:
To do this, apply the product rule:
Let [tex]\(u_1 = x^5\)[/tex] and [tex]\(u_2 = \cos^3 x\)[/tex].
- Differentiate [tex]\(u_1 = x^5\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(x^5) = 5x^4 \][/tex]
- Differentiate [tex]\(u_2 = \cos^3 x\)[/tex] using the chain rule:
[tex]\[ \frac{d}{dx}(\cos^3 x) = 3\cos^2 x \cdot (-\sin x) = -3\cos^2 x \sin x \][/tex]
Therefore:
[tex]\[ \frac{d}{dx}(x^5 \cos^3 x) = (5x^4 \cos^3 x) + (x^5 \cdot -3\cos^2 x \sin x) = 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \][/tex]
Differentiate [tex]\(v = \frac{1}{\sqrt{\sin x}}\)[/tex]:
Rewrite [tex]\(v\)[/tex] in a more convenient form for differentiation:
[tex]\[ v = (\sin x)^{-1/2} \][/tex]
Using the power rule and the chain rule:
[tex]\[ \frac{d}{dx}[(\sin x)^{-1/2}] = -\frac{1}{2} (\sin x)^{-3/2} \cdot \cos x = -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \][/tex]
3. Combine the results using the product rule:
Apply the product rule:
[tex]\[ \frac{d}{dx} y = \frac{d}{dx} \left( \frac{x^5 \cos^3 x}{\sqrt{\sin x}} \right) = \left( 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \right) \cdot \frac{1}{\sqrt{\sin x}} + x^5 \cos^3 x \cdot \left( -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \right) \][/tex]
Simplify each term separately:
- First term:
[tex]\[ \left( 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \right) \cdot \frac{1}{\sqrt{\sin x}} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - \frac{3x^5 \cos^2 x \sin x}{\sqrt{\sin x}} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - 3x^5 \cos^2 x \sqrt{\sin x} \][/tex]
- Second term:
[tex]\[ x^5 \cos^3 x \cdot \left( -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \right) = -\frac{1}{2} \cdot \frac{x^5 \cos^4 x}{(\sin x)^{3/2}} \][/tex]
Combine the terms:
[tex]\[ \frac{dy}{dx} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - 3x^5 \cos^2 x \sqrt{\sin x} - \frac{1}{2} \cdot \frac{x^5 \cos^4 x}{(\sin x)^{3/2}} \][/tex]
Rearranging and combining the terms appropriately yields:
[tex]\[ \frac{dy}{dx} = -\frac{3 x^{5} \sin (x) \cos ^{2}(x)}{\sqrt{\csc (x)}}+\frac{x^{5} \cos ^{3}(x) \cot (x)}{2 \sqrt{\csc (x)}}+\frac{5 x^{4} \cos ^{3}(x)}{\sqrt{\csc (x)}} \][/tex]
Thus, the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ \frac{dy}{dx} = -\frac{3 x^{5} \sin (x) \cos ^{2}(x)}{\sqrt{\csc (x)}}+\frac{x^{5} \cos ^{3}(x) \cot (x)}{2 \sqrt{\csc (x)}}+\frac{5 x^{4} \cos ^{3}(x)}{\sqrt{\csc (x)}} \][/tex]
1. Rewrite the function in a more convenient form:
Given:
[tex]\[ y = \frac{x^5 \cos^3 x}{\sqrt{\csc x}} \][/tex]
Recall that [tex]\(\csc x = \frac{1}{\sin x}\)[/tex], so [tex]\(\sqrt{\csc x} = \sqrt{\frac{1}{\sin x}} = \frac{1}{\sqrt{\sin x}}\)[/tex].
Therefore, the function can be rewritten as:
[tex]\[ y = x^5 \cos^3 x \cdot \frac{1}{\sqrt{\sin x}} = \frac{x^5 \cos^3 x}{\sqrt{\sin x}} \][/tex]
2. Use the product rule and the chain rule:
To differentiate this expression, we will utilize the product rule:
[tex]\[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \][/tex]
where [tex]\(u = x^5 \cos^3 x\)[/tex] and [tex]\(v = \frac{1}{\sqrt{\sin x}}\)[/tex].
Let’s differentiate each part separately:
Differentiate [tex]\(u = x^5 \cos^3 x\)[/tex]:
To do this, apply the product rule:
Let [tex]\(u_1 = x^5\)[/tex] and [tex]\(u_2 = \cos^3 x\)[/tex].
- Differentiate [tex]\(u_1 = x^5\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(x^5) = 5x^4 \][/tex]
- Differentiate [tex]\(u_2 = \cos^3 x\)[/tex] using the chain rule:
[tex]\[ \frac{d}{dx}(\cos^3 x) = 3\cos^2 x \cdot (-\sin x) = -3\cos^2 x \sin x \][/tex]
Therefore:
[tex]\[ \frac{d}{dx}(x^5 \cos^3 x) = (5x^4 \cos^3 x) + (x^5 \cdot -3\cos^2 x \sin x) = 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \][/tex]
Differentiate [tex]\(v = \frac{1}{\sqrt{\sin x}}\)[/tex]:
Rewrite [tex]\(v\)[/tex] in a more convenient form for differentiation:
[tex]\[ v = (\sin x)^{-1/2} \][/tex]
Using the power rule and the chain rule:
[tex]\[ \frac{d}{dx}[(\sin x)^{-1/2}] = -\frac{1}{2} (\sin x)^{-3/2} \cdot \cos x = -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \][/tex]
3. Combine the results using the product rule:
Apply the product rule:
[tex]\[ \frac{d}{dx} y = \frac{d}{dx} \left( \frac{x^5 \cos^3 x}{\sqrt{\sin x}} \right) = \left( 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \right) \cdot \frac{1}{\sqrt{\sin x}} + x^5 \cos^3 x \cdot \left( -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \right) \][/tex]
Simplify each term separately:
- First term:
[tex]\[ \left( 5x^4 \cos^3 x - 3x^5 \cos^2 x \sin x \right) \cdot \frac{1}{\sqrt{\sin x}} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - \frac{3x^5 \cos^2 x \sin x}{\sqrt{\sin x}} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - 3x^5 \cos^2 x \sqrt{\sin x} \][/tex]
- Second term:
[tex]\[ x^5 \cos^3 x \cdot \left( -\frac{1}{2} \cdot \frac{\cos x}{(\sin x)^{3/2}} \right) = -\frac{1}{2} \cdot \frac{x^5 \cos^4 x}{(\sin x)^{3/2}} \][/tex]
Combine the terms:
[tex]\[ \frac{dy}{dx} = \frac{5x^4 \cos^3 x}{\sqrt{\sin x}} - 3x^5 \cos^2 x \sqrt{\sin x} - \frac{1}{2} \cdot \frac{x^5 \cos^4 x}{(\sin x)^{3/2}} \][/tex]
Rearranging and combining the terms appropriately yields:
[tex]\[ \frac{dy}{dx} = -\frac{3 x^{5} \sin (x) \cos ^{2}(x)}{\sqrt{\csc (x)}}+\frac{x^{5} \cos ^{3}(x) \cot (x)}{2 \sqrt{\csc (x)}}+\frac{5 x^{4} \cos ^{3}(x)}{\sqrt{\csc (x)}} \][/tex]
Thus, the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ \frac{dy}{dx} = -\frac{3 x^{5} \sin (x) \cos ^{2}(x)}{\sqrt{\csc (x)}}+\frac{x^{5} \cos ^{3}(x) \cot (x)}{2 \sqrt{\csc (x)}}+\frac{5 x^{4} \cos ^{3}(x)}{\sqrt{\csc (x)}} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
