Example Question - quotient rule
Here are examples of questions we've helped users solve.
Finding the Derivative of a Rational Function
<p>Дана функция \( f(x) = \frac{x^2 - 4}{2x^2 + 4} \). Чтобы найти производную, используем правило дифференцирования частного:</p> <p>Пусть \( u(x) = x^2 - 4 \) и \( v(x) = 2x^2 + 4 \), тогда \( f(x) = \frac{u(x)}{v(x)} \).</p> <p>Производная \( u \) по \( x \): \( u'(x) = 2x \).</p> <p>Производная \( v \) по \( x \): \( v'(x) = 4x \).</p> <p>Теперь используем правило дифференцирования частного \( (u/v)' = \frac{u'v - uv'}{v^2} \):</p> <p>\[ f'(x) = \frac{(2x)(2x^2 + 4) - (x^2 - 4)(4x)}{(2x^2 + 4)^2} \]</p> <p>Упрощаем выражение:</p> <p>\[ f'(x) = \frac{(4x^3 + 8x) - (4x^3 - 16x)}{(2x^2 + 4)^2} \]</p> <p>\[ f'(x) = \frac{4x^3 + 8x - 4x^3 + 16x}{(2x^2 + 4)^2} \]</p> <p>\[ f'(x) = \frac{24x}{(2x^2 + 4)^2} \]</p> <p>Можно дополнительно упростить, вынеся константу за скобки:</p> <p>\[ f'(x) = \frac{24x}{4(x^2 + 2)^2} \]</p> <p>\[ f'(x) = \frac{6x}{(x^2 + 2)^2} \]</p> <p>Это и есть производная функции \( f(x) \).</p>
Calculation of a Function's Derivative
Дадената функция е: <p>\( y = \frac{x^5 - 4}{x^2 - 4x} \)</p> За да намерим производната \( y' \), ще приложим правилото за частно (quotient rule): <p>\( y' = \frac{(u'v - uv')}{v^2} \)</p> където \( u = x^5 - 4 \) и \( v = x^2 - 4x \). След това намираме \( u' \) и \( v' \): <p>\( u' = 5x^4 \)</p> <p>\( v' = 2x - 4 \)</p> Сега изчисляваме производната \( y' \) чрез заместване на \( u', u, v', v \): <p>\( y' = \frac{(5x^4)(x^2 - 4x) - (x^5 - 4)(2x - 4)}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{5x^6 - 20x^5 - (2x^6 - 4x^5 - 8x + 16)}{x^4 - 8x^3 + 16x^2} \)</p> <p>\( y' = \frac{5x^6 - 20x^5 - 2x^6 + 4x^5 + 8x - 16}{x^4 - 8x^3 + 16x^2} \)</p> <p>\( y' = \frac{3x^6 - 16x^5 + 8x - 16}{x^4 - 8x^3 + 16x^2} \)</p> Това е производната на функцията.
Calculating the Derivative of a Rational Function
Пусть функция \( y \) задана как \( y = \frac{x^5 - 4}{x^2 - 4x} \). Тогда ее производная \( y' \) может быть найдена с использованием правила дифференцирования частного: <p>\( y' = \frac{(x^2 - 4x)'(x^5 - 4) - (x^5 - 4)'(x^2 - 4x)}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{(2x - 4)(x^5 - 4) - (5x^4)(x^2 - 4x)}{(x^2 - 4x)^2} \)</p> Теперь раскроем скобки в числителе: <p>\( y' = \frac{2x^6 - 8x - 4x^5 + 16 - 5x^6 + 20x^5}{(x^2 - 4x)^2} \)</p> <p>\( y' = \frac{-3x^6 + 12x^5 - 8x + 16}{(x^2 - 4x)^2} \)</p> Это и будет искомая производная функции \( y \).
Simplifying Logarithmic Expressions
The expression provided in the image, \(\log\left(\frac{a^3b}{c^2}\right)\), can be simplified using the properties of logarithms: The Quotient Rule: \(\log(x/y) = \log(x) - \log(y)\) The Power Rule: \(\log(x^k) = k \cdot \log(x)\) So let's break down the expression step by step: \(\log\left(\frac{a^3b}{c^2}\right)\) = \( \log(a^3b) - \log(c^2) \) (using the Quotient Rule) = \( \log(a^3) + \log(b) - \log(c^2) \) (using the Product Rule: \(\log(xy) = \log(x) + \log(y)\)) = \( 3\cdot\log(a) + \log(b) - 2\cdot\log(c) \) (using the Power Rule) Thus, the correct answer to the expression given in the image is: \(\log(a) + 3\log(b) - 2\log(c)\) This corresponds to answer choice D in the image.
Solving Logarithmic Expressions with Log Rules
To solve the expression given in the image, \( \log_{c^2}(\frac{a^5b}{c^2}) \), we can apply log rules (quotient rule, power rule, and change of base formula). The quotient rule states that \( \log(\frac{x}{y}) = \log(x) - \log(y) \). The power rule states that \( \log(x^a) = a \log(x) \). The change of base formula states that \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \). Let's break down the expression by using these rules: 1. Apply the quotient rule to the argument of the log. \( \log_{c^2}(a^5b) - \log_{c^2}(c^2) \) 2. Apply the power rule to \( a^5 \) and \( c^2 \) within the log expressions. \( 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 \log_{c^2}(c) \) 3. Notice that \( \log_{c^2}(c^2) = 1 \) because the log base and the argument are the same value raised to the same power. Hence, the expression becomes: \( 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 \) 4. Apply the change of base formula to the two remaining log terms. Since we want to express everything in terms of the base c, we get: \( 5 \frac{\log_{c}(a)}{\log_{c}(c^2)} + \frac{\log_{c}(b)}{\log_{c}(c^2)} - 2 \) 5. Simplify by recognizing that \( \log_{c}(c^2) = 2 \). \( 5 \frac{\log_{c}(a)}{2} + \frac{\log_{c}(b)}{2} - 2 \) 6. Simplify the fractions. \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \) 7. Looking at the answer choices given, we can see that our derived expression matches choice D when we distribute the 2 outside of the log: \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \cdot 1 \) \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \log_{c}(c) \) Since \( \log_{c}(c) = 1 \), the 2 just becomes \( 2 \log_{c}(c) \), and thus the answer is: \( \frac{5}{2} \cdot \log_{c}(a) + \frac{1}{2} \cdot \log_{c}(b) - 2 \log_{c}(c) \) So, the correct answer is D: \( 5 \log_{c}(a) + \log_{c}(b) - 2 \log_{c}(c) \)
Derivative of a Function with Respect to x using Quotient Rule and Chain Rule
The image shows a mathematical problem asking to find the derivative with respect to \(x\) of the function \(f(x,y)\), which is provided as: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) \] To find this derivative, we will use the quotient rule along with the chain rule because the function has both \(x\) and \(y\) variables, where \(y\) could be a function of \(x\). The quotient rule is: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] where \(u = 2xy\) and \(v = \sqrt{x^2 + y^2}\). The derivative of \(u = 2xy\) with respect to \(x\) is \(u' = 2y + 2xy'\) (using the product rule and assuming \(y\) is a function of \(x\)). The derivative of \(v = \sqrt{x^2 + y^2}\) with respect to \(x\) is \(v' = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x + 2yy')\) (using the chain rule and assuming \(y\) is a function of \(x\)). Simplifying gives \(v' = \frac{x + yy'}{\sqrt{x^2 + y^2}}\). Now, apply the quotient rule: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) = \frac{(2y + 2xy')\sqrt{x^2 + y^2} - 2xy\left(\frac{x + yy'}{\sqrt{x^2 + y^2}}\right)}{x^2 + y^2} \] Simplify the expression: \[ = \frac{2y(x^2 + y^2) + 2xy'y\sqrt{x^2 + y^2} - 2x^2y - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] Now, combine like terms and simplify further: \[ = \frac{2y^3 + 2xy'y\sqrt{x^2 + y^2} - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] If \(y\) is not a function of \(x\), \(y'\) would simply be 0, and the expression would be further simplified. However, the derivative expressed as it stands takes into consideration that \(y\) may be a function of \(x\). If you have more specific information about \(y\), such as whether it is a constant or a function of \(x\), you can simplify the expression accordingly.
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.
Finding the Derivative of sin(x)/x^2 using Quotient Rule
To find the derivative \(\frac{dy}{dx}\) of the given function \(y = \frac{\sin x}{x^2}\), you can use the quotient rule. The quotient rule states that if you have a function in the form of \(\frac{f(x)}{g(x)}\), then its derivative is given by \[ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] Let \(f(x) = \sin x\) and \(g(x) = x^2\). We need to find the derivatives \(f'(x)\) and \(g'(x)\). \(f'(x) = \cos x\) (since the derivative of \(\sin x\) with respect to \(x\) is \(\cos x\)) \(g'(x) = 2x\) (since the derivative of \(x^2\) with respect to \(x\) is \(2x\)) Now apply the quotient rule: \[ \frac{dy}{dx} = \frac{(\cos x)(x^2) - (\sin x)(2x)}{(x^2)^2} \] Simplifying the expression: \[ \frac{dy}{dx} = \frac{x^2\cos x - 2x\sin x}{x^4} \] This derivative can further be simplified if necessary by factoring out an \(x\) in the numerator (assuming \(x \neq 0\) since we cannot divide by zero): \[ \frac{dy}{dx} = \frac{x(\cos x - 2\frac{\sin x}{x})}{x^4} \] This simplifies to: \[ \frac{dy}{dx} = \frac{\cos x}{x^3} - 2\frac{\sin x}{x^4} \] So this is the derivative of the function \(y = \frac{\sin x}{x^2}\) with respect to \(x\).
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