Question

If z= x^3y+ 3xy^4 where x= sin 2t and y= cos^2t .

Find dz/dx |t=0 by using chain rule.

Answer 1

$\begin{array}{l} \qquad \begin{array}{l} z=x^{3} y+3 x y^{4}, \quad x=\sin 2 t \\ y=\cos ^{2} t \end{array} \\ d z / d x, t=0 \text { by choin rule } \\ \frac{d z}{d t}=\frac{\partial z}{\partial x} \frac{d x}{d t}+\frac{\partial z}{\partial y} \frac{d y}{d t}= \\ \frac{d z}{d t}=\left(3 x^{2} y+3 y^{4}\right) \cdot 2 \cdot \cos (2 t)+\left(x^{3}+12 x y^{3}\right) \cdot 2 \cdot \cos t \cdot(-\sin t) \\ t=0 \quad \frac{d z}{d t}=\left(3 x^{2} y+3 y^{4}\right) \cdot 2 \cos (2 \cdot 0)+\left(x^{3}+12 x y^{3}\right) \cdot(-2 \cos (0) \cdot \sin (0)) \\ \frac{d z}{d t}=3 x^{2} y+3 y^{4} \cdot 2+0 \\ x=\sin (2 \cdot 0)=0, \quad y=\cos ^{2}(0)=1 \\ \frac{d z}{d t}=3 \cdot(0)^{2} \cdot 1+3 \cdot(1)^{4} \cdot 2=6 \end{array}$