COMPASSLearning

Talk to us Register Login

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Derivatives XI

The slope of the tangent to the curve

(y - x^{5})^{2} = x (1 + x^{2})^{2}

at the point

(1, 3)

is

Answer: 8

Solution

At

x = 1

, we have

(y - 1)^{2} = 1 \cdot 4 = 4

, so

y - 1 = \pm 2

. Since the point is

(1, 3)

, we have

y - 1 = 2

, so

y - x^{5} = x (1 + x^{2})

. Thus

y = x (1 + x^{2}) + x^{5}

. Differentiating:

\frac{d y}{d x} = \frac{1}{2 x} (1 + x^{2}) + x (2 x) + 5 x^{4} = \frac{1 + x ^{2}}{2 x} + 2 x x + 5 x^{4}

. At

x = 1

:

\frac{d y}{d x} = \frac{2}{2} + 2 + 5 = 1 + 2 + 5 = 8

.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer

Practise more: all JEE Advanced PYQs by year and chapter →

More JEE Advanced questions on Derivatives XI

Let $R$ denote the set of all real numbers. Consider the polynomial function $f : R \to R$ defined by …
JEE Advanced 2026View solution →
Consider the functions defined implicitly by the equation $y^{3} - 3 y + x = 0$ on various intervals in the real line.
If $x \in (- \infty, - 2) \cup (2, \infty)$ , the equation implicitly defines a unique real valued differentiable function $y = f (x)$ .
If $x \in (- 2, 2)$ , the equation implicitl …
JEE Advanced 2008View solution →
Consider the function $f : (- \infty, \infty) \to (- \infty, \infty)$ defined by
$f (x) = \frac{x ^{2} - a x + 1}{x ^{2} + a x + 1}, 0 < a < 2.$

Which of the following is true?
JEE Advanced 2008View solution →
$\frac{d ^{2} x}{d y ^{2}}$ equals
JEE Advanced 2007View solution →