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Mathematics 15 Online
OpenStudy (anonymous):

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)

OpenStudy (anonymous):

satellite this one:S

OpenStudy (anonymous):

you come up with the doozies

OpenStudy (anonymous):

take the derivative wrt x and get \[1+y'=\frac{1}{x^4+y^{13}}\times (4x^3+13y^{12}y')\]

OpenStudy (anonymous):

can someone please help me

OpenStudy (anonymous):

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'

OpenStudy (anonymous):

sorry my comp froze ehnc ei asked for help again

OpenStudy (anonymous):

\[1+y'=1\times 4\]

OpenStudy (anonymous):

\[y'=3\]

OpenStudy (anonymous):

so it looked bad but substitution made it simple

OpenStudy (anonymous):

hope it is clear, gotta run

OpenStudy (anonymous):

wait so y=3 is the equation?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

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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