Answer :
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]
