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Question: How do you differentiate \(y = \sin (xy)\)?...

How do you differentiate y=sin(xy)y = \sin (xy)?

Explanation

Solution

To differentiate the above equation we need to use implicit differentiation where two variables x and y are differentiated with respect to one of the variables and also applying chain rule and product rule since the expression given above is a composite function.

Complete step by step answer:
The above equation is a composite function which means it contains a function inside another function.
Suppose f(x) and g(x) are two different functions. So, the composite of this will be f(g(x)).
In the above case, sin(x) is one function and xy is another function, therefore sin(xy) is composite.
So, now by applying implicit differentiation,
y=sin(xy) dydx=ddxsin(xy)  \Rightarrow y = \sin (xy) \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\sin (xy) \\\
Further by using chain rule on sin function i.e., ddxsinx=cosududx\dfrac{d}{{dx}}\sin x = \cos u\dfrac{{du}}{{dx}} where u stands for xy.
Therefore, we can write this as,
dydx=cos(xy)(ddx(xy))\Rightarrow \dfrac{{dy}}{{dx}} = \cos (xy)\left( {\dfrac{d}{{dx}}(xy)} \right)
By applying product rule i.e.,
d(uv)dx=udvdx+vdudx\Rightarrow \dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}
On the function xy. Here the first term is multiplied by the derivative of second and second term is multiplied by the derivative of first.

dydx=cos(xy)[x(ddxy+yddxx)] dydx=cos(xy)(xdydx+y(1))  \Rightarrow \dfrac{{dy}}{{dx}} = \cos (xy)\left[ {x\left( {\dfrac{d}{{dx}}y + y\dfrac{d}{{dx}}x} \right)} \right] \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \cos (xy)\left( {x\dfrac{{dy}}{{dx}} + y(1)} \right) \\\

dx when differentiated with dx gives 1. After that open the brackets and multiply cos(xy)\cos (xy) with the terms which were inside the bracket.
dydx=xcos(xy)dydx+ycos(xy)\Rightarrow \dfrac{{dy}}{{dx}} = x\cos (xy)\dfrac{{dy}}{{dx}} + y\cos (xy)
Shifting the term on the left-hand side of the equal sign and taking dydx\dfrac{{dy}}{{dx}} common from the equation.

dydxxcos(xy)dydx=ycos(xy) dydx(1xcos(xy))=ycos(xy)  \Rightarrow \dfrac{{dy}}{{dx}} - x\cos (xy)\dfrac{{dy}}{{dx}} = y\cos (xy) \\\ \Rightarrow \dfrac{{dy}}{{dx}}(1 - x\cos (xy)) = y\cos (xy) \\\

By shifting the term on the right-hand side by dividing it we will get the value for dydx\dfrac{{dy}}{{dx}}
dydx=ycos(xy)(1xcos(xy))\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\cos (xy)}}{{(1 - x\cos (xy))}}
Therefore, by implicit differentiation we get the above solution.

Note: For solving any equation first you need to check whether it is composite or not. If it is composite then focus should be in identifying which is the inner function and which is the outer function. Further it will be easy to apply chain rule and product rule.