Find the dy/dx xy=(x+y)^4

Mathematics

1 answer:

JulijaS [17]4 years ago

8 0

Answer:

dy/dx = (4 (x + y)^3 - y) / ( x - 4(x + y)^3).

Step-by-step explanation:

xy = (x + y)^4

Using implicit differentiation:

x dy/dx + y*1 = 4 (x + y)^3 * (1 + dy/dx)

x dy/dx + y = 4 (x + y)^3 + 4 dy/dx (x + y)^3

x dy/dx - 4 dy/dx (x + y)^3 = 4 (x + y)^3 - y

dy/dx( x - 4(x + y)^3) = 4 (x + y)^3 - y

dy/dx = (4 (x + y)^3 - y) / ( x - 4(x + y)^3)

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What is the equation of the line (in slope-intercept form) that passes through the given point and is

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$\bf x+y=6\implies y = \stackrel{\stackrel{m}{\downarrow }}{-1}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{1}\implies 1}}$

so we're really looking for the equation of a line whose slope is 1 and runs through (-5,-6).

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-6})~\hspace{10em} \stackrel{slope}{m}\implies 1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{1}[x-\stackrel{x_1}{(-5)}] \\\\\\ y+6=1(x+5)\implies y+6=x+5\implies y=x-1$