The slope of the tangent to the curve `(y-x^5)^2=x(1+x^2)^2`
at the point `(1,3)`
is.
Text Solution
Verified by Experts
`(y-x^(5))^(2)=x(1+x^(2))^(2)` Differentiating both sides w.r.t. x, we get `2(y-x^(5))((dy)/(dx)-5x^(4))=1(1+x^(2))^(2)+(x)(2(1+x^(2))(2x))` On putting x=1 , y =3 in above equation, we get `(dy)/(dx)=8`
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