x3y−2x2+y4=8 then dydx at P(1,2) is

A
$\frac{2}{33}$
B
$\frac{- 2}{33}$
C
$\frac{4}{15}$
D
$\frac{- 4}{15}$

Solution:

$\begin{array}{l} x^{3} y - 2 x^{2} + y^{4} = 8 \\ \Rightarrow x^{3} \frac{d y}{d x} + y (3 x^{2}) - 2 (2 x) + 4 y^{3} \frac{d y}{d x} = 0 \\ \Rightarrow x^{3} \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x} = 4 x - 3 x^{2} y \\ \Rightarrow (x^{3} + 4 y^{3}) \frac{d y}{d x} = 4 x - 3 x^{2} y \\ \Rightarrow \frac{d y}{d x} = \frac{4 x - 3 x^{2} y}{x^{3} + 4 y^{3}} \\ \Rightarrow {(\frac{d y}{d x})}_{P (1, 2)} = \frac{4 (1) - 3 {(1)}^{2} (2)}{{(1)}^{3} + 4 {(2)}^{3}} \\ = \frac{4 - 6}{1 + 32} \\ = \frac{- 2}{33} \end{array}$

$x^{3} y - 2 x^{2} + y^{4} = 8$ then $\frac{d y}{d x}$ at $P (1, 2)$ is

Solution:

Related content