Question

Find the equation of tangent to the curve $y=\{\begin{array}{cc}{x}^2\sin \frac{1}{x}& x\ne 0\\ 0& x=0\end{array}$ at $(0,0)$.

Answer 1

here the curve $y=x^2\sin \frac{1}{x},x\ne 0$

& now at $x=0\Rightarrow y=0$

Now, the tangent will be given as

$y=mx+c$

where $m=\frac{dy}{dx}{]}_{(0,0)}$ or $m=\frac{dy}{dx}$

so, $y=x^2\sin \frac{1}{x}$

differentiate $y$ wrt $x$ we get

$\frac{dy}{dx}=2x\sin \frac{1}{2}+x^2\left(\cos \frac{1}{x}\right)\left(\frac{-1}{x^2}\right)$

so, $\frac{dy}{dx}=2\sin \frac{1}{x}-\cos \frac{1}{x}$

now, $y=mx+c$

$y=(2x\sin \frac{1}{x}-\cos \frac{1}{x})x+c$

$y=2x^2\sin \frac{1}{x}-\cos \frac{1}{x}+c$

$y=2y-x\cos \frac{1}{x}+c$

so, $y=x\cos \frac{1}{x}-c$

now, at $(0,0)0=0-c\Rightarrow c=0$

$y=x\cos \frac{1}{x}$ i.e., tangent of the curve $y$