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if x=sin(x+y), find dy/dx

Nov 20, 2018

#1
+24089
+12

if x=sin(x+y), find dy/dx

$$\begin{array}{|rcl|} \hline &&\text{Differentiate both sides of the equation, getting: } \\ D(x) &=& D(\sin(x+y)) \\ &&\text{Use the chain rule}: \\ D(x) &=& \cos(x+y)\cdot D(x+y) \quad | \quad D(x) = 1 \\ 1 &=& \cos(x+y)\cdot D(x+y) \\ 1 &=& \cos(x+y)\cdot(1+y') \\ \dfrac{1}{\cos(x+y)} &=& 1+y' \\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \dfrac{1}{\cos(x+y) }-1 } \\\\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \sec(x+y)-1 } \\ \hline \end{array}$$

Nov 20, 2018
edited by heureka  Nov 20, 2018

#1
+24089
+12
$$\begin{array}{|rcl|} \hline &&\text{Differentiate both sides of the equation, getting: } \\ D(x) &=& D(\sin(x+y)) \\ &&\text{Use the chain rule}: \\ D(x) &=& \cos(x+y)\cdot D(x+y) \quad | \quad D(x) = 1 \\ 1 &=& \cos(x+y)\cdot D(x+y) \\ 1 &=& \cos(x+y)\cdot(1+y') \\ \dfrac{1}{\cos(x+y)} &=& 1+y' \\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \dfrac{1}{\cos(x+y) }-1 } \\\\ \mathbf{ y' } & \mathbf{=} & \mathbf{ \sec(x+y)-1 } \\ \hline \end{array}$$