f'(2) = 20
Equation of the tangent line → y = 20x - 16
Further explanation
To solve this problem there are several basic principles in Derivatives that need to be recalled, namely:
\(y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}\)
\(y = \sin x \Rightarrow \frac{dy}{dx} = \cos x\)
\(y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x\)
\(y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'\)
\(y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}\)
\(y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'\)
\(\texttt{where u and v are functions in variable x}\)
\(\texttt{and u' and v' are derivatives of u and v}\)
Let us now tackle the problem !
Given:
\(f(x) = 8x^2 - x^3\)
\(f'(x) = 8(2x^{2-1}) - 3x^{3-1}\)
\(f'(x) = 16x^1 - 3x^2\)
\(f'(x) = 16x - 3x^2\)
\(f'(2) = 16(2) - 3(2)^2\)
\(f'(2) = 32 - 12\)
\(f'(2) = \boxed{20}\)
Equation of the tangent line to the curve
\(y - y_1 = m ( x - x_1 )\)
\(y - 24 = f'(2) ( x - 2 )\)
\(y - 24 = 20 ( x - 2 )\)
\(y - 24 = 20x - 40\)
\(y = 20x - 40 + 24\)
\(\boxed {y = 20x - 16}\)
Learn more
- Implicit Differentiation : https://brainly.com/question/4711711
- Logarithmic Differentiation : https://brainly.com/question/9226310
- Calculus Problem : https://brainly.com/question/11237537
Answer details
Grade: High School
Subject: Mathematics
Chapter: Differentiation
Keywords: Maximum , Minimum , Value , Function , Variable , Derivation , Differentiation
