Question - Derivatives in Calculus
Solution:
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, $$ a = -\frac{5}{3} $$.
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