Answer :
To find the first derivative of the function [tex]\( g(x) = 4 e^x \cot(x^2) \)[/tex], we will use the product rule and the chain rule because the function comprises the product of two functions: [tex]\(4 e^x\)[/tex] and [tex]\(\cot(x^2)\)[/tex].
The product rule states that if [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] are functions of [tex]\(x\)[/tex], then the derivative of their product is given by:
[tex]\[ (uv)' = u'v + uv' \][/tex]
In this case, let:
[tex]\[ u(x) = 4 e^x \][/tex]
[tex]\[ v(x) = \cot(x^2) \][/tex]
We will first find the derivatives of [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] separately.
1. Derivative of [tex]\(u(x) = 4 e^x\)[/tex]:
[tex]\[ u'(x) = 4 e^x \][/tex]
2. Derivative of [tex]\(v(x) = \cot(x^2)\)[/tex]:
To find the derivative of [tex]\(v(x)\)[/tex], we need to use the chain rule. The chain rule states that if a function [tex]\(v(x)\)[/tex] is the composition of two functions [tex]\(f(g(x))\)[/tex], then the derivative is given by:
[tex]\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \][/tex]
Let:
[tex]\[ w = x^2 \][/tex]
[tex]\[ v(x) = \cot(w) \][/tex]
Then,
[tex]\[ v'(w) = -\csc^2(w) \][/tex]
[tex]\[ w'(x) = 2x \][/tex]
Using the chain rule:
[tex]\[ v'(x) = v'(w) \cdot w'(x) = -\csc^2(x^2) \cdot 2x \][/tex]
[tex]\[ v'(x) = -2x \csc^2(x^2) \][/tex]
Now we apply the product rule:
[tex]\[ g'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ g'(x) = (4 e^x)(\cot(x^2)) + (4 e^x)(-2x \csc^2(x^2)) \][/tex]
[tex]\[ g'(x) = 4 e^x \cot(x^2) - 8x e^x \csc^2(x^2) \][/tex]
Combining these terms, we get:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - 2x \csc^2(x^2) \right) e^x \][/tex]
Therefore, the first derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - \frac{2x}{\sin^2(x^2)} \right) e^x \][/tex]
Or, equivalently:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - 2x \csc^2(x^2) \right) e^x \][/tex]
Thus, the simplified form of the first derivative is:
[tex]\[ g'(x) = 4 \left( -2x \csc^2(x^2) + \cot(x^2) \right) e^x \][/tex]
[tex]\[ g'(x) = 4 \left( -2x \left(\frac{1}{\sin(x^2)}\right)^2 + \cot(x^2) \right) e^x \][/tex]
Therefore, the detailed step-by-step solution for the first derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = 4 \left(-\frac{2x}{\sin^2(x^2)} + \cot(x^2) \right) e^x \][/tex]
The product rule states that if [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] are functions of [tex]\(x\)[/tex], then the derivative of their product is given by:
[tex]\[ (uv)' = u'v + uv' \][/tex]
In this case, let:
[tex]\[ u(x) = 4 e^x \][/tex]
[tex]\[ v(x) = \cot(x^2) \][/tex]
We will first find the derivatives of [tex]\(u(x)\)[/tex] and [tex]\(v(x)\)[/tex] separately.
1. Derivative of [tex]\(u(x) = 4 e^x\)[/tex]:
[tex]\[ u'(x) = 4 e^x \][/tex]
2. Derivative of [tex]\(v(x) = \cot(x^2)\)[/tex]:
To find the derivative of [tex]\(v(x)\)[/tex], we need to use the chain rule. The chain rule states that if a function [tex]\(v(x)\)[/tex] is the composition of two functions [tex]\(f(g(x))\)[/tex], then the derivative is given by:
[tex]\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \][/tex]
Let:
[tex]\[ w = x^2 \][/tex]
[tex]\[ v(x) = \cot(w) \][/tex]
Then,
[tex]\[ v'(w) = -\csc^2(w) \][/tex]
[tex]\[ w'(x) = 2x \][/tex]
Using the chain rule:
[tex]\[ v'(x) = v'(w) \cdot w'(x) = -\csc^2(x^2) \cdot 2x \][/tex]
[tex]\[ v'(x) = -2x \csc^2(x^2) \][/tex]
Now we apply the product rule:
[tex]\[ g'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ g'(x) = (4 e^x)(\cot(x^2)) + (4 e^x)(-2x \csc^2(x^2)) \][/tex]
[tex]\[ g'(x) = 4 e^x \cot(x^2) - 8x e^x \csc^2(x^2) \][/tex]
Combining these terms, we get:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - 2x \csc^2(x^2) \right) e^x \][/tex]
Therefore, the first derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - \frac{2x}{\sin^2(x^2)} \right) e^x \][/tex]
Or, equivalently:
[tex]\[ g'(x) = 4 \left( \cot(x^2) - 2x \csc^2(x^2) \right) e^x \][/tex]
Thus, the simplified form of the first derivative is:
[tex]\[ g'(x) = 4 \left( -2x \csc^2(x^2) + \cot(x^2) \right) e^x \][/tex]
[tex]\[ g'(x) = 4 \left( -2x \left(\frac{1}{\sin(x^2)}\right)^2 + \cot(x^2) \right) e^x \][/tex]
Therefore, the detailed step-by-step solution for the first derivative of [tex]\( g(x) \)[/tex] is:
[tex]\[ g'(x) = 4 \left(-\frac{2x}{\sin^2(x^2)} + \cot(x^2) \right) e^x \][/tex]
