Derivatives in Calculus
The image shows three mathematical problems labeled a), b), and c), all of which are related to calculus. I'll assist you with each one separately.
a) Given \( f(x) = 2x^2 - 5x + 1 \), you are asked to find \( f'(x) \), which is the first derivative of the function with respect to x.
To find \( f'(x) \), differentiate \( f(x) \) term by term:
\[
f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(5x) + \frac{d}{dx}(1)
\]
Using the power rule \( \frac{d}{dx}x^n = nx^{n-1} \) and the constant rule \( \frac{d}{dx}c = 0 \) for a constant c, we get:
\[
f'(x) = 2 \cdot 2x^{2-1} - 5 \cdot 1x^{1-1}
\]
Simplify the expression:
\[
f'(x) = 4x - 5
\]
So, the first derivative \( f'(x) \) of the function \( f(x) = 2x^2 - 5x + 1 \) is \( f'(x) = 4x - 5 \).
b) Next, you are asked to find the derivative \( \frac{dy}{dx} \) given \( y = 3(x^2 - x)^2 \).
We'll use the chain rule here: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), where \( u = x^2 - x \), hence \( y = 3u^2 \).
First, we differentiate \( y \) with respect to \( u \):
\[
\frac{dy}{du} = 3 \cdot 2u^{2-1} = 6u
\]
Then, we differentiate \( u \) with respect to \( x \):
\[
\frac{du}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) = 2x - 1
\]
Now, multiply both derivatives together to get \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = 6u(2x - 1)
\]
Substitute \( u \) back with \( x^2 - x \):
\[
\frac{dy}{dx} = 6(x^2 - x)(2x - 1)
\]
So, \( \frac{dy}{dx} = 6(x^2 - x)(2x - 1) \) is the derivative of \( y = 3(x^2 - x)^2 \) with respect to \( x \).
c) Finally, you are given \( f(x) = ax^2 + 3x - 1 \) and asked to find the value of \( a \), given that the limit \( \lim_{x \to -3} \frac{f(x) - f(-3)}{x - (-3)} = 13 \).
This limit represents the derivative of \( f(x) \) at \( x = -3 \). Let's first find the derivative \( f'(x) \):
\[
f'(x) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(3x) - \frac{d}{dx}(1)
\]
\[
f'(x) = 2ax + 3
\]
Now, evaluate \( f'(x) \) at \( x = -3 \):
\[
f'(-3) = 2a(-3) + 3 = -6a + 3
\]
We are given that \( f'(-3) = 13 \), so:
\[
-6a + 3 = 13
\]
Solving for \( a \):
\[
-6a = 13 - 3
\]
\[
-6a = 10
\]
\[
a = -\frac{10}{6}
\]
\[
a = -\frac{5}{3}
\]
Therefore, \( a = -\frac{5}{3} \).
