$\mathbf{Question:}$ Find the real numbers $a$ and $b$ such that the following function is differentiable at $x=0$

$$ f(x)= \begin{cases} x^{2}+1 &x≥0\\ a\sin x+b\cos x & x<0\\ \end{cases} $$

$\mathbf{My\ attempt:}$

$$ \begin{align} \lim_{x\to 0-}f(x) & =\lim_{x\to 0-}a\sin x +b\cos x \\ & = a\sin (0) + b\cos (0) = b \end{align} $$

$$ \begin{align} \lim_{x \to 0+}f(x) & = \lim_{x \to0+}x^{2}+1 =1 \end{align} $$

So if $f(x)$ is continuous, $\lim_{x \to0-}f(x) = \lim_{x \to0+}f(x)=b$

Therefore, $b=1$

To find $a$, we can set $f'(0)=f'(0)$:

$$ \begin{align} \frac {d}{dx}[x^{2}+1]&=\frac {d}{dx}[a\sin x +b\cos x]\\ 2x&=a\cos x-b\sin x\\ 0&=a(1)+b(0)\\ a&=0\\ \end{align} $$ Therefore, $a=0$

Thus, $ f(x)= \begin{cases} x^{2}+1 &x≥0\\ \cos x & x<0\\ \end{cases} $ is differentiable at $x=0$

  • 1
    $\begingroup$ Looks good to me. Ultimately, in dealing with this sort of problem, one has to use two facts. First, $f(x)$ must be continuous, as any differentiable function is continuous. Like you have observed, this allows us to solve for $b.$ But also, the derivative of $f(x)$ at $0$ must be well-defined, i.e., $\lim_{x \to 0^-} f'(x) = \lim_{x \to 0^+} f'(x).$ Like you have observed, this allows us to solve for $a.$ Good work. $\endgroup$ May 30, 2020 at 16:57

1 Answer 1


This is exactly what one should not do while dealing with differentiability. Instead always use the definition of derivative.

The left derivative at $0$ is given by $$f_{-}'(0)=\lim_{x\to 0^{-}}\frac{f(x)-f(0)}{x}=\lim_{x\to 0^{-}}\frac{a\sin x +b\cos x - 1}{x} =\lim_{x\to 0^{-}}a\frac{\sin x} {x} - \frac{1-\cos x} {x} =a$$ and the right derivative of $f$ at $0$ is given by $$f_{+}' (0)=\lim_{x\to 0^{+}}\frac{f(x)-f(0)}{x}=\lim_{x\to 0^{+}}\frac{x^2+1-1}{x}=0$$ Since the derivative $f'(0)$ exists we must have these two limits same and therefore $a=0$.

Your approach is not at all about differentiability, but it is rather about continuity of the derivative $f'$. And the reason it works for some problems (like the current one) is because derivatives don't have jump discontinuity. And your technique will simply not work in case the derivative is discontinuous at point under consideration.

Try that approach with $$f(0)=0,f(x)=x^2\sin(1/x),x>0$$ and $$f(x) =-x^2\sin(1/x),x<0$$


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