Example Question - power rule
Here are examples of questions we've helped users solve.
Finding the Derivative of a Given Function
<p>Дана функция: \( f(x) = \frac{1}{x^5} - \frac{11}{3} \sqrt[3]{x} \)</p> <p>Применим правило дифференцирования степенной функции:</p> <p>\( f'(x) = \left( \frac{1}{x^5} \right)' - \left( \frac{11}{3} \sqrt[3]{x} \right)' \)</p> <p>\( f'(x) = (-5) \cdot x^{-5 - 1} - \frac{11}{3} \cdot \frac{1}{3} \cdot x^{\frac{1}{3} - 1} \)</p> <p>\( f'(x) = -5x^{-6} - \frac{11}{9}x^{-\frac{2}{3}} \)</p> <p>Приведем производную к окончательному виду:</p> <p>\( f'(x) = -\frac{5}{x^6} - \frac{11}{9x^{\frac{2}{3}}} \)</p>
Find the Derivative of a Function
<p>\( \frac{d}{dx} \left( x^8 - \frac{1}{x} \right) \)</p> <p>= \( \frac{d}{dx} \left( x^8 \right) - \frac{d}{dx} \left( \frac{1}{x} \right) \)</p> <p>= \( 8x^{8-1} - \left( -x^{-2} \right) \)</p> <p>= \( 8x^7 + x^{-2} \)</p> <p>= \( 8x^7 + \frac{1}{x^2} \)</p>
Find the Derivative of the Function
<p>\( f(x) = x^8 - \frac{1}{x} \)</p> <p>Для того чтобы найти производную данной функции, используем правило дифференцирования степенной функции и правило дифференцирования частного.</p> <p>\( f'(x) = \frac{d}{dx} (x^8) - \frac{d}{dx} \left( \frac{1}{x} \right) \)</p> <p>\( f'(x) = 8x^{8-1} - (-1)x^{-1-1} \)</p> <p>\( f'(x) = 8x^7 + x^{-2} \)</p> <p>Таким образом, производная функции \( f(x) \) равна:</p> <p>\( f'(x) = 8x^7 + x^{-2} \)</p>
Finding the Derivative of a Trigonometric Function Raised to a Power
\[ \begin{align*} f(x) &= \sin^2(x) \\ \frac{d}{dx}f(x) &= \frac{d}{dx}[\sin^2(x)] \\ &= 2\sin(x) \cdot \cos(x) \\ &= \sin(2x) \end{align*} \]
Simplifying Complex Expression with Fractions and Exponents
To solve this expression, we should first simplify each part and then multiply them together. Let's start with the first part of the expression: The first part is a fraction with an exponent of -2: \(\left(\frac{{2p^3}}{{\sqrt[5]{q}}}\right)^{-2}\) Let's express the fifth root as fractional exponent and distribute the -2 exponent to each part inside the brackets: \(= \left(2p^3 \cdot q^{-\frac{1}{5}}\right)^{-2}\) \(= 2^{-2} \cdot \left(p^3\right)^{-2} \cdot \left(q^{-\frac{1}{5}}\right)^{-2}\) Now, apply the power of a power rule by multiplying the exponents: \(= 2^{-2} \cdot p^{-6} \cdot q^{\frac{2}{5}}\) The second part is: \(\frac{4}{p^{-1}q}\) Before multiplying both parts together, let's rewrite everything with positive exponents. For the first part, remember we need to move the terms with negative exponents to the denominator to make the exponents positive: \(= \frac{q^{\frac{2}{5}}}{2^2p^6}\) \(= \frac{q^{\frac{2}{5}}}{4p^6}\) For the second part: \(\frac{4}{p^{-1}q} = \frac{4p}{q}\), because moving \(p^{-1}\) to the numerator makes the exponent positive. Now, we multiply these two expressions together: \(\left(\frac{q^{\frac{2}{5}}}{4p^6}\right) \cdot \left(\frac{4p}{q}\right)\) Multiplying the numerators and the denominators separately: \(\frac{q^{\frac{2}{5}} \cdot 4p}{4p^6 \cdot q}\) Notice that \(4\) in the numerator and denominator cancels out and one p from the numerator cancels with one p from the denominator: \(\frac{q^{\frac{2}{5}} \cdot p}{p^6 \cdot q}\) Split up the exponents for the q terms and subtract the exponents for the p terms: \(= \frac{q^{\frac{2}{5}}}{p^{6-1}q}\) \(= \frac{q^{\frac{2}{5}}}{p^5q}\) Now, subtract the exponents for q, since dividing powers with the same base means you subtract the exponents: \(= \frac{1}{p^5q^{1-\frac{2}{5}}}\) \(= \frac{1}{p^5q^{\frac{5}{5}-\frac{2}{5}}}\) \(= \frac{1}{p^5q^{\frac{3}{5}}}\) And that's our result with positive indices: \[ \frac{1}{p^5q^{\frac{3}{5}}} \]
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.
Derivative of a Quadratic Function using Power Rule
To find the derivative of the function y = x^2 with respect to x, you can use the power rule. The power rule states that for any function y = x^n, where n is a constant, the derivative, dy/dx, is given by n*x^(n-1). Applying the power rule to y = x^2: dy/dx = 2*x^(2-1) = 2*x So, the derivative of y = x^2 is dy/dx = 2x.
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 Calculation of a Function
La imagen muestra una pregunta de cálculo sobre derivadas. La función dada es \( f(x) = x^{2/3} \), y se nos pide determinar \( f'(x) \), es decir, la derivada primera de \( f(x) \). Para resolver este problema, utilizaremos la regla de potencias para la derivación. La regla dice que la derivada de \( x^n \) es \( n \cdot x^{n-1} \). Veamos cómo aplicar esto a nuestra función. Tenemos \( f(x) = x^{2/3} \). Entonces, utilizando la regla de potencias: \( f'(x) = \frac{2}{3} \cdot x^{\frac{2}{3} - 1} \) \( f'(x) = \frac{2}{3} \cdot x^{-\frac{1}{3}} \) \( f'(x) = \frac{2}{3} \cdot \frac{1}{x^{\frac{1}{3}}} \) \( f'(x) = \frac{2}{3x^{\frac{1}{3}}} \) Por lo tanto, la derivada de \( f(x) = x^{2/3} \) es \( f'(x) = \frac{2}{3x^{1/3}} \), que en las opciones correspondientes a la pregunta sería la opción B.
Differentiation of Mathematical Functions
The image is blurry, but it seems to show four mathematical functions that need to be differentiated with respect to \( x \). I'll attempt to identify them and differentiate each one: (a) \( y = (x + n)^1 \) Using the power rule, we get: \( \frac{dy}{dx} = 1 \cdot (x + n)^{1 - 1} \cdot \frac{d}{dx}(x + n) \) \( \frac{dy}{dx} = (x + n)^0 \cdot \frac{d}{dx}(x + n) \) Since \( (x + n)^0 = 1 \) and \( \frac{d}{dx}(x + n) = 1 \), then \( \frac{dy}{dx} = 1 \cdot 1 = 1 \) (b) \( f(x) = (2x + n)^1 \) Applying the power rule: \( f'(x) = 1 \cdot (2x + n)^{1 - 1} \cdot \frac{d}{dx}(2x + n) \) \( f'(x) = (2x + n)^0 \cdot \frac{d}{dx}(2x + n) \) Since \( (2x + n)^0 = 1 \) and \( \frac{d}{dx}(2x + n) = 2 \), then \( f'(x) = 1 \cdot 2 = 2 \) (c) \( y = (3 - 4x)^5 \) Using the chain rule, we get: \( \frac{dy}{dx} = 5(3 - 4x)^{4}(-4) \) \( \frac{dy}{dx} = -20(3 - 4x)^{4} \) (d) \( g(x) = (3z - 4x)^2 \) (There appears to be a typo in the variable used in the original function. Assuming it's supposed to be \( x \), not \( z \), and differentiating accordingly:) \( g'(x) = 2(3 - 4x)^1(-4) \) \( g'(x) = -8(3 - 4x) \) If the variables \( n \) or \( z \) are constants, then my differentiation is correct. If \( n \) or \( z \) are not constants, and you meant a different variable or power, please provide the correct expressions.
Derivative of Product of Functions
The given functions are \( f(x) = x^3 + 1 \) and \( g(x) = x^2 \). We need to find the derivative of the product of these functions, \( f(x) \cdot g(x) \), at the point \( x = -1 \). To solve this, we first find the product of the functions: \[ f(x) \cdot g(x) = (x^3 + 1)(x^2) = x^5 + x^2 \] Now we take the derivative of the product: \[ \frac{d}{dx}(f(x) \cdot g(x)) = \frac{d}{dx}(x^5 + x^2) \] Using the power rule for derivatives \( \frac{d}{dx}(x^n) = nx^{n-1} \), the derivatives of each term are: \[ \frac{d}{dx}(x^5) = 5x^4 \] \[ \frac{d}{dx}(x^2) = 2x \] Combine them: \[ \frac{d}{dx}(f(x) \cdot g(x)) = 5x^4 + 2x \] Finally, we evaluate this derivative at \( x = -1 \): \[ \frac{d}{dx}(f(x) \cdot g(x))|_{x = -1} = 5(-1)^4 + 2(-1) = 5 - 2 = 3 \] Therefore, the derivative of the function \( f(x) \cdot g(x) \) at the point \( x = -1 \) is 3.
Calculating Derivative of a Function and Finding its Value
The image shows a function f(x) = 2x^7 - 5x + 1 and asks to calculate f'(x) given f(x) = 1. To find f'(x), we need to differentiate f(x) with respect to x. Taking the derivative term by term, we get: f'(x) = d/dx (2x^7) - d/dx (5x) + d/dx (1) Using the power rule for differentiation, which states that d/dx (x^n) = n*x^(n-1), we get: f'(x) = 7 * 2x^(7-1) - 5 * 1x^(1-1) f'(x) = 14x^6 - 5 Now we have an expression for f'(x). However, the problem states that this should be evaluated given f(x) = 1. This means we have to find x such that 2x^7 - 5x + 1 = 1. From the equation, we can simplify to: 2x^7 - 5x = 0 Removing the constant term from both sides since it does not impact finding x for which f(x) = 1, we need an additional step to find x that satisfies this equation. Generally, this would require solving a 7th-degree polynomial, which may or may not be solvable in terms of radicals. However, the task does not specify needing to find the value of x that makes f(x) = 1. Instead, if you need to find f'(1), then you would substitute x with 1: f'(1) = 14(1)^6 - 5 f'(1) = 14 - 5 f'(1) = 9 So, if you're required to find f'(1), the answer is 9. If you have to find f'(x) for the value of x such that f(x) = 1, the determination of x from the given equation 2x^7 - 5x + 1 = 1 will be more complex and may not have a straightforward solution.
Solving Indefinite Integral of Polynomial Expression
The image shows the integral of a polynomial expression. I'll guide you through the steps to solve the indefinite integral: ∫(8x^3 - x^2 + 5x - 1)dx To integrate this polynomial term by term, you would use the power rule for integration. The power rule states that the integral of x^n dx is (1/(n+1)) * x^(n+1) + C, where C is the constant of integration. Applying this rule to each term, you would get: ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (1/4) * x^4 = 2x^4 ∫(-x^2) dx = (-1) * ∫x^2 dx = (-1) * (1/3) * x^3 = -1/3x^3 ∫5x dx = 5 * ∫x dx = 5 * (1/2) * x^2 = 5/2x^2 ∫(-1) dx = -1 * ∫dx = -1 * x = -x Adding them all together, you get: 2x^4 - 1/3x^3 + 5/2x^2 - x + C Therefore, the integral of the given expression is: 2x^4 - 1/3x^3 + 5/2x^2 - x + C
Integrating Polynomials Using Power Rule
To solve the integral provided in the image: ∫(8x^3 - x^2 + 5x - 1)dx You need to integrate each term separately using the power rule for integration. The power rule states that the integral of x^n with respect to x is (x^(n+1))/(n+1), where n ≠ -1. Applying the power rule to each term: For 8x^3, ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (x^(3+1))/(3+1) = 8 * (x^4)/4 = 2x^4 For -x^2, ∫(-x^2) dx = - ∫x^2 dx = - (x^(2+1))/(2+1) = - (x^3)/3 For 5x, ∫5x dx = 5 * ∫x dx = 5 * (x^(1+1))/(1+1) = 5 * (x^2)/2 = (5/2)x^2 For -1, ∫(-1) dx = - ∫1 dx = -x Now, combine all the integrated terms: ∫(8x^3 - x^2 + 5x - 1)dx = 2x^4 - (x^3)/3 + (5/2)x^2 - x + C Where C is the constant of integration.
Evaluating Definite Integrals of Polynomial Functions using Power Rule
This is a definite integral of a polynomial function, which we can evaluate using the Power Rule for integration. The Power Rule states that the integral of x^n is (x^(n+1)) / (n+1) plus a constant (C), for any real number n ≠ -1. So let's evaluate your integral: ∫(8x^3 - x^2 + 5x - 1) dx Integration term by term: ∫8x^3 dx = 8 * ∫x^3 dx = 8 * (x^(3+1) / (3+1)) = 2 * x^4 ∫-x^2 dx = - ∫x^2 dx = - (x^(2+1) / (2+1)) = - (1/3) * x^3 ∫5x dx = 5 * ∫x dx = 5 * (x^(1+1) / (1+1)) = (5/2) * x^2 ∫-1 dx = -x Combining these results: 2x^4 - (1/3)x^3 + (5/2)x^2 - x Without limits of integration given, we cannot evaluate for specific numbers. If there were limits, you would substitute the upper limit into the result for x, then subtract the result from substituting the lower limit. Since no limits are provided, this is the indefinite integral result, and don't forget to add the constant of integration C: 2x^4 - (1/3)x^3 + (5/2)x^2 - x + C
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