Respuesta :
This is also written as sec^2(x)
Using chain rule, we can separate this into 2sec(x)*d/dx(sec(x))
Using trig derivative identities, we know that d/dx(sec(x))=tan(x)sec(x)
Therefore, we have 2sec(x)*tan(x)sec(x), which simplifies down to 2tan(x)sec^2(x).
Hope this helps!
Using chain rule, we can separate this into 2sec(x)*d/dx(sec(x))
Using trig derivative identities, we know that d/dx(sec(x))=tan(x)sec(x)
Therefore, we have 2sec(x)*tan(x)sec(x), which simplifies down to 2tan(x)sec^2(x).
Hope this helps!
Answer:
[tex]\displaystyle \frac{dy}{dx} = 2 \tan (x) \sec^2 (x)[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y = \sec^2 (x)[/tex]
Step 2: Differentiate
- Basic Power Rule [Derivative Rule - Chain Rule]: [tex]\displaystyle y' = 2 \sec (x) \cdot \frac{d}{dx}[\sec (x)][/tex]
- Trigonometric Differentiation: [tex]\displaystyle y' = 2 \sec (x) \cdot \sec (x) \tan (x)[/tex]
- Simplify: [tex]\displaystyle y' = 2 \sec^2 (x) \tan (x)[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation