Question

exercises 3.8 implicit differentiation
score: 12/20 answered: 6/10
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question 7
0/2 pts 100 99 details
given the equation below, find \\( \frac { d y } { d x } \\).
\\( - 11 x ^ { 7 } + 2 x ^ { 22 } y + y ^ { 9 } = - 8 \\)
hint: you will need to use the product rule on the middle
term.
\\( \frac { d y } { d x } = \\)
now, find the equation of the tangent line to the curve at
\\( ( 1,1 ) \\). write your answer in \\( m x + b \\) format
\\( y = \\)
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Explanation:

Step1: Differentiate each term with respect to \(x\)

• For \(-11x^{7}\), using the power rule \(\frac{d}{dx}(x^n)=nx^{n - 1}\), we get \(\frac{d}{dx}(-11x^{7})=-77x^{6}\).
• For \(2x^{22}y\), using the product rule \((uv)^\prime = u^\prime v+uv^\prime\) where \(u = 2x^{22}\) and \(v=y\). Then \(u^\prime=44x^{21}\) and \(v^\prime=\frac{dy}{dx}\), so \(\frac{d}{dx}(2x^{22}y)=44x^{21}y + 2x^{22}\frac{dy}{dx}\).
• For \(y^{9}\), using the chain rule \(\frac{d}{dx}(y^n)=ny^{n - 1}\frac{dy}{dx}\), we get \(\frac{d}{dx}(y^{9})=9y^{8}\frac{dy}{dx}\).
• The derivative of the constant \(-8\) is \(0\).

So, \(-77x^{6}+44x^{21}y + 2x^{22}\frac{dy}{dx}+9y^{8}\frac{dy}{dx}=0\).

Step2: Solve for \(\frac{dy}{dx}\)

Group the terms with \(\frac{dy}{dx}\):
\((2x^{22}+9y^{8})\frac{dy}{dx}=77x^{6}-44x^{21}y\)
Then \(\frac{dy}{dx}=\frac{77x^{6}-44x^{21}y}{2x^{22}+9y^{8}}\).

Step3: Find the slope \(m\) at the point \((1,1)\)

Substitute \(x = 1\) and \(y = 1\) into \(\frac{dy}{dx}\):
\(m=\frac{77(1)^{6}-44(1)^{21}(1)}{2(1)^{22}+9(1)^{8}}=\frac{77 - 44}{2+9}=\frac{33}{11}=3\).

Step4: Find the equation of the tangent line

Using the point - slope form \(y - y_1=m(x - x_1)\) with \(x_1 = 1\), \(y_1 = 1\) and \(m = 3\).
\(y-1=3(x - 1)\)
Expand: \(y-1=3x-3\)
Rearrange to \(y=mx + b\) form: \(y=3x-2\).

Answer:

\(\frac{dy}{dx}=\frac{77x^{6}-44x^{21}y}{2x^{22}+9y^{8}}\); \(y = 3x-2\)