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JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2014 · Paper 1 · Mathematics · Application of Derivatives

The slope of the tangent to the curve (y - x ^ 5 ) ^ 2 = x (1 + x ^...

The slope of the tangent to the curve (y−x5){\left(y - {x}^{5}\right)}^{2} = \text{x} {\left(1 + {\text{x}}^{2}\right)}^{2} at the point (1, 3) is ________.

Correct answer

0

Step-by-step solution

22\left(y - {x}^{5}\right)\left(\dfrac{dy}{dx} - 5{x}^{4}\right)<br><br >= 1{\left(1 + {x}^{2}\right)}^{2} + \left(x\right)\left(2\left(1 + {x}^{2}\right)\left(2x\right)\right)
Now put x=1,y=3x = 1, y = 3 and dydx=m.\dfrac{dy}{dx} = m .
22\left(3 - 1\right)\left(m - 5\right) = 1\left(4\right) + \left(1\right)\left(4\right)\left(2\right)<br><br >m - 5 = \dfrac{12}{4}<br><br >m = 5 + 3 = 8<br><br >\dfrac{dy}{dx} = m = 8 .
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