Determine The Derivative Of Ax^2
Determine The Derivative Ofax X2
Determine the derivative of: a(x) = x2 cos x2. Given the function f(x) = 23 x3 + 132 x2 – 45x + 171, solve the equation f′(x) = 0. Determine the first and second derivatives of g(x) = 3x3 – 5x10 – 1x5.
Paper For Above instruction
Introduction
Calculus is foundational to understanding the behavior of functions, particularly through the concept of derivatives. Derivatives provide insights into the rate of change, slope of tangent lines, and optimization problems. This paper focuses on computing derivatives for specific functions and solving equations involving derivatives, illustrating core calculus techniques such as the product rule, chain rule, and polynomial differentiation.
Derivative of a(x) = x^2 cos x^2
The first task involves finding the derivative of \( a(x) = x^2 \cos x^2 \). This function combines polynomial and trigonometric components through multiplication, necessitating the product rule. The product rule states that for functions \( u(x) \) and \( v(x) \), the derivative of \( u v \) is \( u' v + u v' \).
Here, let:
- \( u(x) = x^2 \)
- \( v(x) = \cos x^2 \)
Compute:
- \( u'(x) = 2x \)
- \( v'(x) \) involves the chain rule since \( v(x) \) is composite:
\( v'(x) = -\sin x^2 \times 2x = -2x \sin x^2 \)
Applying the product rule:
\[
a'(x) = u'(x) v(x) + u(x) v'(x) = 2x \cos x^2 + x^2 \times (-2x \sin x^2) = 2x \cos x^2 - 2x^3 \sin x^2
\]
This derivative encompasses the combination of polynomial and trigonometric functions, illustrating the use of the product and chain rules.
Solving \( f'(x) = 0 \) for \( f(x) = 23x^3 + 132x^2 - 45x + 171 \)
The second task involves finding the critical points by setting the derivative of \( f(x) \) to zero and solving for \( x \).
First, differentiate \( f(x) \):
\[
f'(x) = \frac{d}{dx} (23x^3 + 132x^2 - 45x + 171) = 69x^2 + 264x - 45
\]
Set \( f'(x) = 0 \):
\[
69x^2 + 264x - 45 = 0
\]
Divide through by 3 for simplicity:
\[
23x^2 + 88x - 15 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 23 \), \( b = 88 \), and \( c = -15 \):
\[
x = \frac{-88 \pm \sqrt{88^2 - 4 \times 23 \times (-15)}}{2 \times 23}
\]
\[
x = \frac{-88 \pm \sqrt{7744 + 1380}}{46} = \frac{-88 \pm \sqrt{9124}}{46}
\]
Calculating the square root:
\[
\sqrt{9124} \approx 95.55
\]
Thus, the critical points are:
\[
x \approx \frac{-88 \pm 95.55}{46}
\]
Compute both solutions:
- \( x \approx \frac{-88 + 95.55}{46} = \frac{7.55}{46} \approx 0.164 \)
- \( x \approx \frac{-88 - 95.55}{46} = \frac{-183.55}{46} \approx -3.991 \)
These critical points indicate potential local maxima or minima, which can be further analyzed with the second derivative test.
First and Second Derivatives of \( g(x) = 3x^3 - 5x^{10} - x^5 \)
The third task involves differentiating \( g(x) = 3x^3 - 5x^{10} - x^5 \), and then finding its second derivative.
First derivative \( g'(x) \):
Differentiate each term:
\[
g'(x) = \frac{d}{dx} (3x^3) - 5 \frac{d}{dx} (x^{10}) - \frac{d}{dx} (x^5) = 9x^2 - 50x^9 - 5x^4
\]
Second derivative \( g''(x) \):
Differentiate \( g'(x) \):
\[
g''(x) = \frac{d}{dx} (9x^2 - 50x^9 - 5x^4) = 18x - 450x^8 - 20x^3
\]
The second derivative informs about concavity and inflection points, central concepts in analyzing the shape and behavior of the function.
Conclusion
Derivatives are essential tools in calculus for analyzing the behavior of functions. Complex functions involving products and compositions require careful application of the product rule and chain rule, as demonstrated with \( a(x) = x^2 \cos x^2 \). Solving derivative equations like \( f'(x) = 0 \) helps determine critical points that are vital in optimization problems. Calculating higher derivatives, such as in \( g(x) \), reveals insights into the function's concavity and inflection points. Mastery of these techniques is fundamental for advanced calculus applications in science, engineering, and economics.
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