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 ^{2} = \text{x} {\left(1 + {\text{x}}^{2}\right)}^{2} at the point (1, 3) is ________.
Correct answer
0
Step-by-step solution
\left(y - {x}^{5}\right)\left(\dfrac{dy}{dx} - 5{x}^{4}\right)= 1{\left(1 + {x}^{2}\right)}^{2} + \left(x\right)\left(2\left(1 + {x}^{2}\right)\left(2x\right)\right)
Now put and
\left(3 - 1\right)\left(m - 5\right) = 1\left(4\right) + \left(1\right)\left(4\right)\left(2\right)m - 5 = \dfrac{12}{4}m = 5 + 3 = 8\dfrac{dy}{dx} = m = 8 .
Now put and
\left(3 - 1\right)\left(m - 5\right) = 1\left(4\right) + \left(1\right)\left(4\right)\left(2\right)m - 5 = \dfrac{12}{4}m = 5 + 3 = 8\dfrac{dy}{dx} = m = 8 .