Question

homework5: problem 5
(1 point)
use implicit differentiation to find the slope of the tangent line to the curve defined by ( 5 x y ^ { 9 } + 6 x y = 11 ) at the point ( ( 1,1 ) ).
the slope of the tangent line to the curve at the given point is ( \frac { - 11 } { 15 } ).
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Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(5xy^{9}+6xy = 11\) with respect to \(x\) using the product rule \((uv)^\prime=u^\prime v + uv^\prime\).
For the term \(5xy^{9}\), let \(u = 5x\) and \(v=y^{9}\). Then \((5xy^{9})^\prime=5y^{9}+5x\times9y^{8}y^\prime=5y^{9}+45xy^{8}y^\prime\).
For the term \(6xy\), let \(u = 6x\) and \(v = y\). Then \((6xy)^\prime=6y+6xy^\prime\).
The derivative of the right - hand side \(11\) is \(0\). So, \((5xy^{9}+6xy)^\prime=(11)^\prime\) gives \(5y^{9}+45xy^{8}y^\prime+6y + 6xy^\prime=0\).

Step2: Solve for \(y^\prime\)

Factor out \(y^\prime\): \(y^\prime(45xy^{8}+6x)=-(5y^{9}+6y)\). Then \(y^\prime=\frac{-(5y^{9}+6y)}{45xy^{8}+6x}\).

Step3: Substitute \(x = 1\) and \(y = 1\)

Substitute \(x = 1\) and \(y = 1\) into the formula for \(y^\prime\).
\(y^\prime=\frac{-(5\times1^{9}+6\times1)}{45\times1\times1^{8}+6\times1}=\frac{-(5 + 6)}{45+6}=\frac{-11}{51}\).

Answer:

\(\frac{-11}{51}\)