<s>[INST] Find the values $a$ and $b$ such that the function is differentiable at $x=0$ $\mathbf{Question:}$
Find the values $a$ and $b$ such that the function is differentiable at $x=0$
$$
f(x)=
\begin{cases}
x^{2}+1 &x≥0\\
a\sin x+b\cos x & x<0\\
\end{cases}
$$
$\mathbf{Solution:}$
$f(x)$ is differentiable at $x=0$ if $f'(0)$ exists. This implies that for $f$ to be differentiable at $x=0$, the left hand limit and the right hand limit must exist and be equal.
$$
\begin{align}
\lim_{x\to 0-}f'(0) & =\lim_{x\to 0-}\frac{f(x)-f(0)}{x} \\
& =\lim_{x\to 0-}\frac{a\sin x +b\cos x-1}{x} \\
& =\lim_{x\to 0-}\frac{a\sin x}{x}+\frac{b\cos x-1}{x}=a \\
\end{align}
$$
$$
\begin{align}
\lim_{x\to 0+}f'(0) & =\lim_{x\to 0+}\frac{f(x)-f(0)}{x} \\
& =\lim_{x\to 0+}\frac{x^{2}+1-1}{x} = 0
\end{align}
$$
Therefore, $a=0$
To find $b$, we can use the fact that if $f(x)$ is differentiable at $x=0$ then, it must be continuous at $x=0$. 
So if $f(x)$ is continuous, $\lim_{x \to0-}f(x) = \lim_{x \to0+}f(x)=b$
$$
\begin{align}
\lim_{x\to 0-}f(x) & =\lim_{x\to 0-}a\sin x +b\cos x \\
& = a\sin (0) + b\cos (0) = b
\end{align}
$$
$$
\begin{align}
\lim_{x \to 0+}f(x) & = \lim_{x \to0+}x^{2}+1 =1
\end{align}
$$
Therefore, $b=1$
Thus, 
$
f(x)=
\begin{cases}
x^{2}+1 &x≥0\\
\cos x & x<0\\
\end{cases}
$
is differentiable at $x=0$
 [/INST] $$f(0)=1$$
$$\lim_{x\to 0^+}f(x)=\lim_{x\to0^+}(a\sin(x)+b\cos(x))=b$$
$f$ is continuous at $x=0$ if $b=f(0)=1$.
$$f'(0^-)=\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}$$
$$=\lim_{x\to0^-}x=\color{red}{0}$$
$f$ is differentiable at $x=0$ if
$$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\color{red}{0}$$
or
$$\lim_{x\to0^+}\frac{a\sin(x)+\cos(x)-1}{x}=0$$
but
$$\sin(x)\sim x \; and \; \cos(x)-1\sim \frac{-x^2}{2}$$
thus
$a=0$.
</s>