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Use implicit differentiation to find an equation of the tangent line to the graph at the given point. x + y − 1 = ln(x^4 + y^13), (1, 0)
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satellite this one:S
you come up with the doozies
take the derivative wrt x and get \[1+y'=\frac{1}{x^4+y^{13}}\times (4x^3+13y^{12}y')\]
can someone please help me
now solve this for y' but since all you really need is the derivative when x = 1 and y = 0, just replace x by 1 and y by 0 in that equation and solve for y'
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sorry my comp froze ehnc ei asked for help again
\[1+y'=1\times 4\]
\[y'=3\]
so it looked bad but substitution made it simple
hope it is clear, gotta run
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wait so y=3 is the equation?
?
The slope of the tangent would be 3. So the tangent line is:\[y-0=3(x-1)\]Giving:\[y=3x-3\]
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