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To find the derivative of the function [tex]\( y(x) = \cos^2 (\sqrt{1 - x^2}) + \cos (\sqrt{1 - x^2}) \)[/tex], we will apply the chain rule, product rule, and the derivatives of basic functions.
1. Function Definition:
[tex]\[ y(x) = \cos^2 (\sqrt{1 - x^2}) + \cos (\sqrt{1 - x^2}) \][/tex]
2. Chain Rule:
Let's first differentiate [tex]\( \cos (\sqrt{1 - x^2}) \)[/tex]:
[tex]\[ u = \sqrt{1 - x^2} \quad \text{and} \quad v = \cos(u) \][/tex]
The derivative of [tex]\( v \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ \frac{dv}{du} = -\sin(u) \][/tex]
The derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot \frac{d}{dx} (1 - x^2) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \][/tex]
Combining these using the chain rule:
[tex]\[ \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
3. Differentiating the Entire Function:
Now apply this to the entire function. Let's break it into two parts:
[tex]\( f_1(x) = \cos^2 (\sqrt{1 - x^2}) \)[/tex]
and
[tex]\( f_2(x) = \cos (\sqrt{1 - x^2}) \)[/tex]
- First part [tex]\( f_1(x) = (\cos (\sqrt{1 - x^2}))^2 \)[/tex]:
Use [tex]\(z = \cos (\sqrt{1 - x^2}) \)[/tex] and apply the chain rule:
[tex]\[ \frac{d}{dx} (z^2) = 2z \cdot \frac{dz}{dx} \][/tex]
where
[tex]\[ \frac{dz}{dx} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
Therefore,
[tex]\[ \frac{d}{dx} (\cos^2 (\sqrt{1 - x^2})) = 2 \cos (\sqrt{1 - x^2}) \cdot \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
- Second part [tex]\( f_2(x) = \cos (\sqrt{1 - x^2}) \)[/tex]:
We have already differentiated this:
[tex]\[ \frac{d}{dx} (\cos (\sqrt{1 - x^2})) = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
4. Combine the Derivatives:
Combine the results for [tex]\( f_1(x) \)[/tex] and [tex]\( f_2(x) \)[/tex]:
[tex]\[ \frac{dy}{dx} = 2 \cos (\sqrt{1 - x^2}) \cdot \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
Simplify further if possible:
[tex]\[ \frac{dy}{dx} = \frac{2 x \cos (\sqrt{1 - x^2}) \sin (\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin (\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \left( 2 \cos (\sqrt{1 - x^2}) + 1 \right) \][/tex]
Thus, the derivative of the given function [tex]\( y(x) \)[/tex] is:
[tex]\[ \boxed{\frac{dy}{dx} = \frac{2x \sin(\sqrt{1 - x^2}) \cos(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}}} \][/tex]
1. Function Definition:
[tex]\[ y(x) = \cos^2 (\sqrt{1 - x^2}) + \cos (\sqrt{1 - x^2}) \][/tex]
2. Chain Rule:
Let's first differentiate [tex]\( \cos (\sqrt{1 - x^2}) \)[/tex]:
[tex]\[ u = \sqrt{1 - x^2} \quad \text{and} \quad v = \cos(u) \][/tex]
The derivative of [tex]\( v \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[ \frac{dv}{du} = -\sin(u) \][/tex]
The derivative of [tex]\( u \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} (\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot \frac{d}{dx} (1 - x^2) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \][/tex]
Combining these using the chain rule:
[tex]\[ \frac{dv}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} = -\sin(u) \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
3. Differentiating the Entire Function:
Now apply this to the entire function. Let's break it into two parts:
[tex]\( f_1(x) = \cos^2 (\sqrt{1 - x^2}) \)[/tex]
and
[tex]\( f_2(x) = \cos (\sqrt{1 - x^2}) \)[/tex]
- First part [tex]\( f_1(x) = (\cos (\sqrt{1 - x^2}))^2 \)[/tex]:
Use [tex]\(z = \cos (\sqrt{1 - x^2}) \)[/tex] and apply the chain rule:
[tex]\[ \frac{d}{dx} (z^2) = 2z \cdot \frac{dz}{dx} \][/tex]
where
[tex]\[ \frac{dz}{dx} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
Therefore,
[tex]\[ \frac{d}{dx} (\cos^2 (\sqrt{1 - x^2})) = 2 \cos (\sqrt{1 - x^2}) \cdot \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
- Second part [tex]\( f_2(x) = \cos (\sqrt{1 - x^2}) \)[/tex]:
We have already differentiated this:
[tex]\[ \frac{d}{dx} (\cos (\sqrt{1 - x^2})) = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
4. Combine the Derivatives:
Combine the results for [tex]\( f_1(x) \)[/tex] and [tex]\( f_2(x) \)[/tex]:
[tex]\[ \frac{dy}{dx} = 2 \cos (\sqrt{1 - x^2}) \cdot \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
Simplify further if possible:
[tex]\[ \frac{dy}{dx} = \frac{2 x \cos (\sqrt{1 - x^2}) \sin (\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin (\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} \left( 2 \cos (\sqrt{1 - x^2}) + 1 \right) \][/tex]
Thus, the derivative of the given function [tex]\( y(x) \)[/tex] is:
[tex]\[ \boxed{\frac{dy}{dx} = \frac{2x \sin(\sqrt{1 - x^2}) \cos(\sqrt{1 - x^2})}{\sqrt{1 - x^2}} + \frac{x \sin(\sqrt{1 - x^2})}{\sqrt{1 - x^2}}} \][/tex]
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