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if7y−2y5−2x+3x3−25=0,finddydxat(2,1). Hence, find the equation of the line tangent to the curve at this point?
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use implcicit differentiation
\[\frac{dy}{dx}=y^{'}\] \[7y^{'}-2(5)y^{'}y^4-2+9x^2=0\]
\[\frac{ dy }{dx }(7-10y ^{4})=2-9x ^{2}\]
yes
\[dydx=\frac{ 2−9x2 }{ 7−10y4 } ???\]
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yes did you understand the procedure on how to use implicit diff
from here put (2,1)
\[\frac{ 2-9(2)^{2} }{ 7-10(1)^{4} }=\frac{ 34 }{ 3 }\]
yes that's your gradient m=34/3 so nw use y-y1=m(x-x1)
\[y=\frac{ 34 }{ 3 }x-\frac{ 68 }{ 3 }+1= \frac{ 34 }{ 3 }x-\frac{ 65 }{ 3 }\]
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