Answer :
To differentiate the function [tex]\( y = \sec(\theta) \tan(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex], we can use the product rule. The product rule states that if we have a function [tex]\( y \)[/tex] which is the product of two functions [tex]\( u(\theta) \)[/tex] and [tex]\( v(\theta) \)[/tex], then the derivative of [tex]\( y \)[/tex] with respect to [tex]\( \theta \)[/tex] is given by:
[tex]\[ \frac{d}{d\theta}[u(\theta)v(\theta)] = u'(\theta)v(\theta) + u(\theta)v'(\theta) \][/tex]
For our function:
[tex]\[ y = \sec(\theta) \tan(\theta) \][/tex]
we will let
[tex]\[ u(\theta) = \sec(\theta) \quad \text{and} \quad v(\theta) = \tan(\theta) \][/tex]
First, let's find the derivatives [tex]\( u'(\theta) \)[/tex] and [tex]\( v'(\theta) \)[/tex].
1. The derivative of [tex]\( u(\theta) = \sec(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ u'(\theta) = \frac{d}{d\theta} \sec(\theta) = \sec(\theta) \tan(\theta) \][/tex]
2. The derivative of [tex]\( v(\theta) = \tan(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ v'(\theta) = \frac{d}{d\theta} \tan(\theta) = \sec^2(\theta) \][/tex]
Now, applying the product rule:
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = \sec(\theta) \tan(\theta) \cdot \tan(\theta) + \sec(\theta) \cdot \sec^2(\theta) \][/tex]
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = \sec(\theta) \tan^2(\theta) + \sec^3(\theta) \][/tex]
Next, let's express [tex]\( \sec^3(\theta) \)[/tex] in a way that incorporates simpler terms:
[tex]\[ \sec^3(\theta) = \sec(\theta) \cdot \sec^2(\theta) \][/tex]
Since [tex]\( \sec^2(\theta) = 1 + \tan^2(\theta) \)[/tex] (based on the trigonometric identity),
[tex]\[ \sec^3(\theta) = \sec(\theta) (1 + \tan^2(\theta)) \][/tex]
Therefore,
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) \left( 1 + \tan^2(\theta) \right) \][/tex]
Combine the terms:
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) + \sec(\theta) \tan^2(\theta) \][/tex]
Collecting like terms, we get:
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) \tan^2(\theta) + \sec(\theta) \][/tex]
Combining the common factors:
[tex]\[ (2 \tan^2(\theta) + 1) \sec(\theta) \][/tex]
Thus, the derivative of [tex]\( y = \sec(\theta) \tan(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex] is:
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = (\tan^2(\theta) + 1) \sec(\theta) + \tan^2(\theta) \sec(\theta) \][/tex]
[tex]\[ \frac{d}{d\theta}[u(\theta)v(\theta)] = u'(\theta)v(\theta) + u(\theta)v'(\theta) \][/tex]
For our function:
[tex]\[ y = \sec(\theta) \tan(\theta) \][/tex]
we will let
[tex]\[ u(\theta) = \sec(\theta) \quad \text{and} \quad v(\theta) = \tan(\theta) \][/tex]
First, let's find the derivatives [tex]\( u'(\theta) \)[/tex] and [tex]\( v'(\theta) \)[/tex].
1. The derivative of [tex]\( u(\theta) = \sec(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ u'(\theta) = \frac{d}{d\theta} \sec(\theta) = \sec(\theta) \tan(\theta) \][/tex]
2. The derivative of [tex]\( v(\theta) = \tan(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex]:
[tex]\[ v'(\theta) = \frac{d}{d\theta} \tan(\theta) = \sec^2(\theta) \][/tex]
Now, applying the product rule:
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = \sec(\theta) \tan(\theta) \cdot \tan(\theta) + \sec(\theta) \cdot \sec^2(\theta) \][/tex]
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = \sec(\theta) \tan^2(\theta) + \sec^3(\theta) \][/tex]
Next, let's express [tex]\( \sec^3(\theta) \)[/tex] in a way that incorporates simpler terms:
[tex]\[ \sec^3(\theta) = \sec(\theta) \cdot \sec^2(\theta) \][/tex]
Since [tex]\( \sec^2(\theta) = 1 + \tan^2(\theta) \)[/tex] (based on the trigonometric identity),
[tex]\[ \sec^3(\theta) = \sec(\theta) (1 + \tan^2(\theta)) \][/tex]
Therefore,
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) \left( 1 + \tan^2(\theta) \right) \][/tex]
Combine the terms:
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) + \sec(\theta) \tan^2(\theta) \][/tex]
Collecting like terms, we get:
[tex]\[ \sec(\theta) \tan^2(\theta) + \sec(\theta) \tan^2(\theta) + \sec(\theta) \][/tex]
Combining the common factors:
[tex]\[ (2 \tan^2(\theta) + 1) \sec(\theta) \][/tex]
Thus, the derivative of [tex]\( y = \sec(\theta) \tan(\theta) \)[/tex] with respect to [tex]\( \theta \)[/tex] is:
[tex]\[ \frac{d}{d\theta} \left[ \sec(\theta) \tan(\theta) \right] = (\tan^2(\theta) + 1) \sec(\theta) + \tan^2(\theta) \sec(\theta) \][/tex]
