Example Question - product rule
Here are examples of questions we've helped users solve.
Finding the Derivative of a Function Involving a Product of a Polynomial and a Trigonometric Function
Для нахождения производной данной функции \( f(x) = (4x^4 + 2) \cos(x) \), воспользуемся правилом произведения. <p> \( f'(x) = (4x^4 + 2)' \cos(x) + (4x^4 + 2) \cdot (\cos(x))' \) </p> <p> \( f'(x) = (16x^3) \cos(x) + (4x^4 + 2) \cdot (-\sin(x)) \) </p> <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p> Итак, производная функции \( f(x) \): <p> \( f'(x) = 16x^3 \cos(x) - (4x^4 + 2) \sin(x) \) </p>
Solving a Derivative of a Product of Binomial Expressions
<p>To find the derivative of \( y = (7 + x)(7 - x)^2 \), we need to apply the product rule, which states that if \( y = u \cdot v \), then the derivative \( y' = u' \cdot v + u \cdot v' \). Let \( u = 7 + x \) and \( v = (7 - x)^2 \).</p> <p>First, we find the derivative of \( u \) with respect to \( x \): \( u' = \frac{d}{dx}(7 + x) = 0 + 1 = 1 \).</p> <p>Next, we find the derivative of \( v \) with respect to \( x \). For this, we can use the chain rule since \( v \) is a function raised to a power: \( v = (7 - x)^2 \), let \( w = 7 - x \), then \( v = w^2 \).</p> <p>Derivative of \( w \) with respect to \( x \) is: \( w' = \frac{d}{dx}(7 - x) = 0 - 1 = -1 \).</p> <p>Now, apply the chain rule to find the derivative of \( v \): \( v' = 2(w)^{2-1} \cdot w' = 2(7 - x) \cdot -1 = -2(7 - x) \).</p> <p>Finally, applying the product rule: \( y' = u' \cdot v + u \cdot v' = 1 \cdot (7 - x)^2 + (7 + x) \cdot -2(7 - x) \). Simplify the expression to get the final result: \( y' = (7 - x)^2 - 2(7 + x)(7 - x) \).</p> <p>For further simplification, \( y' \) can be expanded by squaring the binomial and applying the distributive property.</p>
Derivative of a Function Involving Square Roots and Chains
The function is \( f(x) = x \sqrt{1 + \sqrt{2x + 6}} \). To find \( f'(x) \), we'll need to use the product rule and the chain rule. The product rule states that if you have a function \( u(x) \cdot v(x) \), then the derivative of this function with respect to \( x \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). The chain rule states that the derivative of a composite function \( u(v(x)) \) is \( u'(v(x)) \cdot v'(x) \). Let \( u(x) = x \) and \( v(x) = \sqrt{1 + \sqrt{2x + 6}} \). Then \( u'(x) = 1 \) since the derivative of \( x \) with respect to \( x \) is 1. For \( v(x) \), let \( h(x) = 1 + \sqrt{2x + 6} \), so \( v(x) = \sqrt{h(x)} \). First, find the derivative of \( h(x) \): \( h'(x) = 0 + \frac{1}{2} (2x + 6)^{-\frac{1}{2}} \cdot (2) = \frac{1}{\sqrt{2x + 6}} \). Next, we use the chain rule to find the derivative of \( v(x) \): \( v'(x) = \frac{1}{2} h(x)^{-\frac{1}{2}} \cdot h'(x) = \frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \). Now we can find \( f'(x) \) using the product rule: \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \) \( f'(x) = 1 \cdot \sqrt{1 + \sqrt{2x + 6}} + x \cdot \left(\frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \right) \) \( f'(x) = \sqrt{1 + \sqrt{2x + 6}} + \frac{x}{2(1 + \sqrt{2x + 6})\sqrt{2x + 6}} \). This is the derivative of the given function. Be sure to simplify where possible to get the final expression for \( f'(x) \).
Derivatives of Mathematical Functions
The given image shows two mathematical functions, f(x) and g(x), and asks for the derivation of these functions at certain points. The functions are: f(x) = x^3 + 1 g(x) = x^2 - x You are asked to do two things: a) Find the derivative of f(x)g(x) at x = 1. b) Find the derivative of the quotient f(x)/g(x) at x = -1. Let's solve each part separately. a) To find the derivative of f(x)g(x), we first need to find the product of these functions and then take the derivative of the product, and evaluate it at x = 1. The product of f(x) and g(x) is: f(x)g(x) = (x^3 + 1)(x^2 - x) Instead of multiplying and then differentiating, we will use the product rule for differentiation, which states that (fg)' = f'g + fg'. Here are the derivatives of f(x) and g(x): f'(x) = d/dx (x^3 + 1) = 3x^2 g'(x) = d/dx (x^2 - x) = 2x - 1 Now apply the product rule: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) = (3x^2)(x^2 - x) + (x^3 + 1)(2x - 1) Now we evaluate this derivative at x = 1: (3(1)^2)(1^2 - 1) + (1^3 + 1)(2(1) - 1) = 3(1)(0) + (2)(1) = 0 + 2 = 2 So the derivative of f(x)g(x) at x = 1 is 2. b) To find the derivative of the quotient f(x)/g(x), we will use the quotient rule for differentiation, which states that (f/g)' = (f'g - fg') / g^2. Again, we will need to use f'(x) = 3x^2 and g'(x) = 2x - 1, and f(x) = x^3 + 1 and g(x) = x^2 - x. Applying the quotient rule, we get: (f(x)/g(x))' = (f'(x)g(x) - f(x)g'(x)) / (g(x))^2 = ((3x^2)(x^2 - x) - (x^3 + 1)(2x - 1)) / (x^2 - x)^2 Now we evaluate this derivative at x = -1: = ((3(-1)^2)((-1)^2 - (-1)) - ((-1)^3 + 1)(2(-1) - 1)) / ((-1)^2 - (-1))^2 = ((3)(1)(1 + 1) - ((-1) + 1)(-2 - 1)) / (1 + 1)^2 = (6 - (0)(-3)) / 4 = 6 / 4 = 3 / 2 So the derivative of f(x)/g(x) at x = -1 is 3/2.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com