Differentiate the given function $y=xsinx$ w.r.t x

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HINT- Proceed the solution of this question using product rule of differentiation as given function is product of two functions.

Complete step-by-step answer:
We have:


Which is the product of two functions, and so we apply the Product Rule to find its Differentiation:
\[\dfrac{{d{\text{(uv)}}}}{{dx}} = {\text{u}}\dfrac{{d{\text{v}}}}{{dx}} + {\text{v}}\dfrac{{d{\text{u}}}}{{dx}}\] or (uv)'= u(dv) +(du)v

So our question is y=xsinx;

Assume u=x, so on differentiate both side w.r.t x
⇒$\dfrac{{d{\text{u}}}}{{dx}} = \dfrac{{d{\text{x}}}}{{dx}}$
⇒$\dfrac{{d{\text{u}}}}{{dx}} = 1$

And v=sinx, so again differentiate both side w.r.t x

⇒$\dfrac{{d{\text{v}}}}{{dx}} = \dfrac{{d\sin {\text{x}}}}{{dx}}$ (Using $\dfrac{{d\sin {\text{x}}}}{{dx}} = \cos {\text{x}}$)
⇒$\dfrac{{d{\text{v}}}}{{dx}} = \cos {\text{x}}$

\[\dfrac{{d{\text{(uv)}}}}{{dx}} = {\text{u}}\dfrac{{d{\text{v}}}}{{dx}} + {\text{v}}\dfrac{{d{\text{u}}}}{{dx}}\]

On putting: u=x, v=sinx, $\dfrac{{d{\text{u}}}}{{dx}} = 1$, $\dfrac{{d{\text{v}}}}{{dx}} = \cos {\text{x}}$

Hence we get,
$\dfrac{{d{\text{x}}{\text{.sinx}}}}{{dx}} = {\text{x(cosx) + (1)sinx}}$

So, $\dfrac{{d{\text{y}}}}{{dx}} = {\text{x}}{\text{.cosx + sinx}}$

Most derivative rules tell us how to differentiate a specific kind of function, like the rule for the derivative of sin(x) or the power rule. However, there are three very important rules that are generally applicable, and depend on the structure of the function we are differentiating. These are the product, quotient, and chain rules, so learn the basic formulas quickly.

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