Find the derivative of a function at a certain point using the difference quotient formula Calculus

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here, we have:

f\left(x\right)= \dfrac{2x^2}{1-x}

Therefore:

For all h\notin\{0{,}1\},

\dfrac{f\left(0+h\right)-f\left(0\right)}{h}=\dfrac{\dfrac{2\left(0+h\right)^2}{1-\left(0+h\right)}-\dfrac{2\left(0\right)^2}{1-\left(0\right)}}{h}= \dfrac{\dfrac{2h^2}{1-h} - 0}{h} = \dfrac{2h}{1-h}

And:

\lim\limits_{h \rightarrow 0} \dfrac{2h}{1-h} = \dfrac{2\left(0\right)}{1-0} = \dfrac{0}{1} = 0

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus

f'\left(a\right) = l

Here, we have:

f\left(x\right) = \dfrac{2x^2 + 1}{3x^2 + 2}

Therefore:

For all h\neq 0,

\dfrac{f\left(0+h\right) - f\left(0\right)}{h} =\dfrac{\dfrac{2\left(0+h\right)^2 + 1}{3\left(0+h\right)^2 + 2} - \dfrac{2\left(0\right)^2 + 1}{3\left(0\right)^2 + 2}}{h} = \dfrac{\dfrac{2h^2 + 1}{3h^2 + 2} - \dfrac{1}{2}}{h} = \dfrac{\dfrac{2\left(2h^2 + 1\right) - \left(3h^2 + 2\right)}{2\left(3h^2 + 2\right)}}{h} = \dfrac{\dfrac{h^2}{2\left(3h^2 + 2\right)}}{h} = \dfrac{h}{2\left(3h^2 + 2\right)}

And:

\lim\limits_{h \rightarrow 0} \dfrac{h}{2\left(3h^2 + 2\right)} = \dfrac{\left(0\right)}{2\left(3\left(0\right)^2 + 2\right)} = \dfrac{0}{4} = 0

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here, we have:

f\left(x\right) = \dfrac{x}{x^2 + 1}

Therefore:

For all h\neq 0,

\dfrac{f\left(1+h\right) - f\left(1\right)}{h} = \dfrac{\dfrac{1 + h}{\left(1+h\right)^2 + 1} - \dfrac{1}{1^2 + 1}}{h} = \dfrac{2\left(h+1\right) - \left(h^2 + 2h + 2\right)}{2h\left(h^2 + 2h + 2\right)}= \dfrac{2h + 2 - h^2 - 2h - 2}{2h\left(h^2 + 2h + 2\right)} = \dfrac{-h}{2\left(h^2 + 2h + 2\right)}

And:

\lim\limits_{h \rightarrow 0} \dfrac{-h}{2\left(h^2 + 2h + 2\right)} = \dfrac{0}{2\left(0 +0 + 2\right)} = 0

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here we have:

f\left(x\right) = 2x^2 + 2x + 6

Therefore:

For all h\neq 0,

\dfrac{f\left(-2+h\right) - f\left(-2\right)}{h} = \dfrac{2\left(-2 + h\right)^2 + 2\left(-2 + h\right) + 6 - \left( 2\left(-2\right)^2 + 2\left(-2\right) + 6 \right)}{h} = \dfrac{2h^2 - 8h + 8 + 2h - 4 + 6 - 8 + 4 - 6}{h} = \dfrac{\left(8 - 8\right) + \left(6 - 6\right) + \left(4 - 4\right) + 2h^2 - 6h}{h} = \dfrac{2h^2 - 6h}{h} = 2h - 6\\

And:

\lim\limits_{h \rightarrow 0} \left( 2h - 6 \right) = 2\left(0\right) - 6 = -6

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here, we have:

f\left(x\right) = x^2 - 4x + 6

Therefore:

For all h\neq 0,

\dfrac{f\left(2+h\right) - f\left(2\right)}{h} = \dfrac{\left( \left(2+h\right)^2 - 4\left(2+h\right) + 6 \right) - \left( \left(2\right)^2 - 4\left(2\right) + 6 \right)}{h} = \dfrac{4 + 4h + h^2 - 8 - 4h + 6 - 4 + 8 - 6}{h} = \dfrac{\left(4 - 4\right) + \left(8 - 8\right) + \left(6 - 6\right) + \left(4h - 4h\right) + h^2}{h} = \dfrac{h^2}{h} = h

And:

\lim\limits_{h \rightarrow 0} h = 0

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here, we have:
f\left(x\right) = \dfrac{\left(x + 2\right)^2 + 3}{x - 3}

Therefore:

For all h\notin\{0{,}6\},

\dfrac{f\left(-3+h\right) - f\left(-3\right)}{h} = \dfrac{\dfrac{\left(-3 + h + 2\right)^2 + 3}{-3 + h - 3} - \dfrac{\left(-3 + 2\right)^2 + 3}{-3 - 3}}{h} = \dfrac{\dfrac{\left(h - 1\right)^2 + 3}{h - 6} - \dfrac{\left(-1\right)^2 + 3}{-6}}{h} = \dfrac{3\left(h^2 - 2h + 4\right) + 2h - 12}{h\left(3h - 18\right)} = \dfrac{3h^2 - 4h}{h\left(3h - 18\right)} = \dfrac{3h - 4}{3h - 18}

And:

\lim\limits_{h \rightarrow 0} \dfrac{3h^2 - 4}{3h - 18} = \dfrac{0 - 4}{0 - 18} = \dfrac{2}{9}

A function f is differentiable at x = a if and only if there is a real number l such that:

\lim\limits_{h \to 0}\dfrac{f\left(a + h\right) - f\left(a\right)}{h}=l

Thus:

f'\left(a\right) = l

Here, we have:

f \left(x\right) = \dfrac{5-x}{10-3x^2}

Therefore:

For all h\notin\left\{0,\sqrt{\dfrac{10}{3}},-\sqrt{\dfrac{10}{3}}\right\},

\dfrac{f\left(0+h\right) - f\left(0\right)}{h} = \dfrac{\dfrac{5-\left(0+h\right)}{10-3\left(0+h\right)^2} - \dfrac{5-0}{10-3\left(0\right)^2}}{h} = \dfrac{\dfrac{5-h}{10-3h^2} - \dfrac{5}{10}}{h} = \dfrac{\left(5-h\right)\left(2\right) - \left(1\right)\left(10-3h^2\right)}{h\left(2\right)\left(10-3h^2\right)} = \dfrac{10 - 2h - 10 + 3h^2}{2h\left(10-3h^2\right)} = \dfrac{3h - 2}{2\left(10-3h^2\right)}

And:

\lim\limits_{h \rightarrow 0} \dfrac{3h - 2}{2\left(10-3h^2\right)} = \dfrac{-2}{2\left(10\right)} = \dfrac{-1}{10}