Calculus Archive: Questions from October 15, 2022
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Compute \( y^{\prime} \) for \( y=\frac{-3}{\sqrt[3]{\left(\sin x+e^{x}\right)^{4}}} \) (A) \( y^{\prime}=\frac{4\left(\cos x+e^{x}\right)}{\sqrt[3]{\left(\sin x+e^{x}\right)^{7}}} \) (B) \( y^{\prime2 answers -
I need solution of 48 and 50
41. \( y=x^{2}-6 x+7 \) 42. \( f(x)=6 x^{2}-x^{3} \) 43. \( f(x)=x(4-x)^{3} \) 44. \( g(x)=(x-1)^{2}(x-3)^{2} \) 45. \( y=x^{2}+\frac{2}{x} \) 46. \( f(x)=\frac{x^{2}}{x-2} \) 47. \( y=x^{2}-32 \sqrt{2 answers -
agnor the first img
if \( y=\frac{\sqrt{x^{3}+x}}{x^{2}+1} \), compute \( y^{\prime} \) A \( y^{\prime}=\frac{\left(x^{2}+1\right)\left(3 x^{2}+1\right)-4 x\left(x^{3}+x\right)}{2\left(\sqrt{x^{3}+x}\right)\left(x^{2}+1\2 answers -
1. Considere \( w=x^{2}-2 x y+y^{2}, x=r+\theta, y=r-\theta \) para determinar \( \frac{\partial w}{\partial r} \& \frac{\partial w}{\partial \theta} \). II. Considere \( w=x y \cos (z), x=t, y=t^{2}0 answers -
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Select the description and the sketch of the domain of the function. \[ \begin{aligned} f(x, y)=& \frac{2 x y}{\left(y-x^{2}\right)} \\ &\left\{(x, y) \mid yx^{2}\right\} \end{aligned} \] \( \left\{(x2 answers -
determines the gradient of the function and the direction of maximum growth of the function at the given point
IV. Determine el gradiente de la función y la dirección de máximo crecimiento de la función en el punto dado. \[ f(x, y)=x \tan (y) ; P\left(2, \frac{\pi}{3}\right) \]2 answers -
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b) \( \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}\left(x^{2}+y^{2}\right)^{3 / 2} d y d x \) II. Establezca y evalúe el integral en las coordenadas más convenientes para determinar el área de la región2 answers -
2. \( \int \sin ^{5} x \cos ^{3} x d x= \) a. \( -\frac{\sin ^{6} x}{6}-\frac{\sin ^{8} x}{8}+C \) b. \( \frac{\cos ^{4} x}{4}-\frac{\cos ^{6} x}{6}+C \) c. \( \frac{\cos ^{2} x}{2}+\frac{\sin ^{4} x}0 answers -
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Find \( y^{\prime} \) and \( y^{\prime \prime} \) by implicit differentiation. \[ \begin{array}{l} x^{2}+x y+y^{2}=4 \\ y^{\prime}=-\frac{2 x+y}{x+2 y} \\ y^{\prime \prime}=-\frac{x^{2}+x y+y^{2}}{2 x2 answers -
what does y" =
Find \( y^{\prime} \) and \( y^{\prime \prime} \) by implicit differentiation. \( 2 x^{3}-3 y^{3}=6 \) \( y^{\prime}=\frac{2 x^{2}}{3 y^{2}} \) \( y^{\prime \prime}=\frac{-4 x^{5}+27 y^{5}}{-6 x^{4} y2 answers -
the error in applying simpson's rule with n=4 on the interval [0,1] to f(x) is
4. El error al aplicar la regla de Simpson con \( n=4 \) en el intervalo \( [0,1] \) a \( f(x)=x^{4}-2 x^{2}+x+2 \) es a. \( \left|E_{S}\right| \leq 0.0005208333 \) b. \( \left|E_{S}\right| \leq 0.0080 answers -
Please help with question 20 & 30
Computing (1st-Order) Partial Derivatives In Exercises 1-78, compute the partial derivatives (i.e., 1st-order partials) of the given functions. 1. \( f(x, y)=x^{3} y^{4}+6 x^{5}-3 y^{3} \) 2. \( f(x,1 answer -
I. Determine la derivada (escoja 3 de los ejercicios) 1) \( f(x)=\arctan \left(e^{2 x}\right) \) 2) \( y=\frac{\operatorname{arcsen}(3 x)}{x} \) 3) \( y=\operatorname{sen}(\arccos (x)) \) II. Trabaje2 answers -
what does y"=
Find \( y^{\prime} \) and \( y^{\prime \prime} \) by implicit differentiation. \[ \begin{array}{c} x^{2}+x y+y^{2}=4 \\ y^{\prime}=-\frac{2 x+y}{x+2 y} \\ y^{\prime \prime}=-\frac{2 x y+6 y^{2}-2 x^{22 answers -
II. Establezca y evalüe el integral en las coordenadas más convenientes para determinar el àrea de la región. A. B.0 answers -
Utilice coordenadas polares para escribir y evaluar la integral doble \( \int_{R} \int f(x, y) d A \) Para \( f(x, y)=x+y \) donde \( R: x^{2}+y^{2} \leq 4, x \geq 0, y \geq 0 \)0 answers -
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#13
Given \( f(x, y)=-6 x^{6}-2 x^{2} y^{3}+y^{2} \) \[ f_{x}(x, y)= \] \[ f_{y}(x, y)= \] \[ f_{x x}(x, y)= \] \[ f_{x y}(x, y)= \]0 answers -
Help :/
I. Considere \( w=x^{2}-2 x y+y^{2}, x=r+\theta, y=r-\theta \) para determinar \( \frac{\partial w}{\partial r} \& \frac{\partial w}{\partial \theta} \). II. Considere \( w=x y \cos (z), x=t, y=t^{2}2 answers -
2 answers
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Given \( f(x, y)=-6 x^{6}-2 x^{2} y^{3}+y^{2} \) \[ f_{x}(x, y)= \] \( f_{y}(x, y)= \) \[ f_{x x}(x, y)= \] \[ f_{x y}(x, y)= \]2 answers -
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I. Evalúe el integral cambiando a coordenadas polares a) \( \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} \cos \left(x^{2}+y^{2}\right) d y d x \) b) \( \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}\left(x^{2}+y^2 answers -
II. Establezca y evalúe el integral en las coordenadas más convenientes para determinar el área de la región.2 answers -
Evaluate the integral \( \iint_{R} \frac{y \sqrt{16-y^{2}}}{\cos ^{2} x} d A \) where \( R=\left\{(x, y): 0 \leqslant x \leqslant \frac{\pi}{4}, 0 \leqslant y \leqslant 4\right\} \)2 answers -
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Thank you in advance whoever it is that works on this!
Evalúe el integral cambiando a coordenadas polares a) \( \int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} \cos \left(x^{2}+y^{2}\right) d y d x \) b) \( \int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}\left(x^{2}+y^{2}0 answers -
Section 6 Solve each DE using the variation of parameters technigy (25) \( y^{\prime \prime}-y=x+3 \) Altempt all (26) \( y^{\prime \prime}-2 y^{\prime}+y=e^{x} \) 3 problems (27) \( x^{\prime \prime}2 answers -
The sum of iterated integrals \[ \int_{-2}^{-1} \int_{-\sqrt{y+2}}^{\sqrt{y+2}} f(x, y) d x d y+\int_{-1}^{2} \int_{y}^{\sqrt{y+2}} f(x, y) d x d y \] is equal: Select one: a. \( \int_{-1}^{2} \int_{x2 answers -
2 answers