Question

Differentiate xsiny with respect to x

Answer 1

We use what's called the product rule for differentiating equation in the form of u.v;

y=u.v$y=u.v$

dydx=u∗dvdx+v∗dudx$\frac{dy}{dx}=u\ast \frac{dv}{dx}+v\ast \frac{du}{dx}$

For this case we are also going to use the fact that;

ddxf(y)=f′(y)dydx$\frac{d}{dx}f(y)=f^{\prime }(y)\frac{dy}{dx}$

This is true by chain rule as you can see below;

u=f(y)$u=f(y)$

dudx=dudy∗dydx$\frac{du}{dx}=\frac{du}{dy}\ast \frac{dy}{dx}$

So;

y=x∗sin(y)$y=x\ast sin(y)$

We let u=x$u=x$ and v=sin(y)$v=sin(y)$

We differentiate both sides with respect to X;

dydx=x∗cos(y)dydx+sin(y)$\frac{dy}{dx}=x\ast cos(y)\frac{dy}{dx}+sin(y)$

We have to make dydx$\frac{dy}{dx}$ the subject so we bring them on one side;

dydx−x∗cos(y)dydx=sin(y)$\frac{dy}{dx}-x\ast cos(y)\frac{dy}{dx}=sin(y)$

Factorising gives;

dydx(1−xcos(y))=sin(y)$\frac{dy}{dx}(1-xcos(y))=sin(y)$

dydx=sin(y)1−xcos(y)