fon = xexpCx) ts
Angles in Circles
Angles within a circle are feasible to create with the help of different properties of the circle such as radii, tangents, and chords. The radius is the distance from the center of the circle to the circumference of the circle. A tangent is a line made perpendicular to the radius through its endpoint placed on the circle as well as the line drawn at right angles to a tangent across the point of contact when the circle passes through the center of the circle. The chord is a line segment with its endpoints on the circle. A secant line or secant is the infinite extension of the chord.
Arcs in Circles
A circular arc is the arc of a circle formed by two distinct points. It is a section or segment of the circumference of a circle. A straight line passing through the center connecting the two distinct ends of the arc is termed a semi-circular arc.
see attached
![The image shows a handwritten mathematical expression on lined paper. The expression reads as follows:
\[ \frac{d}{dx} \left[ x e^{x} \right] \]
This notation represents the derivative of the function \( x e^{x} \) with respect to \( x \).
To find this derivative, apply the product rule of differentiation, which states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:
\[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \]
In this case, let \( u(x) = x \) and \( v(x) = e^{x} \).
1. The derivative of \( u(x) = x \) is \( u'(x) = 1 \).
2. The derivative of \( v(x) = e^{x} \) is \( v'(x) = e^{x} \).
Using the product rule:
\[ \frac{d}{dx} \left[ x e^{x} \right] = x \frac{d}{dx} \left[ e^{x} \right] + e^{x} \frac{d}{dx} \left[ x \right] \]
\[ = x e^{x} + e^{x} \cdot 1 \]
\[ = x e^{x} + e^{x} \]
Therefore, the derivative of \( x e^{x} \) with respect to \( x \) is:
\[ \frac{d}{dx} \left[ x e^{x} \right] = e^{x} (x + 1) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faaa612e4-bb6a-4a7c-9bcb-a4980796bdd1%2F0251739e-f60f-46a1-8de5-0f00394fcb98%2Fqkdlr42_reoriented.jpeg&w=3840&q=75)
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