Why Does My Calculated Derivative Differ from Mathematica's?

[Orion1]

I am attempting to find the derivative of this function using the definition of derivative, however my solution is not the same as Mathematica's solution.

Is my formulation incorrect on some protocol somewhere?

G(t) = \frac{4t}{t + 1}
G'(t) = \lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h}

\lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h} = \lim_{h \rightarrow 0} 4 [\frac{(t + h)(t + 1)}{(t + h + 1)(t + 1)} - \frac{t(t + h + 1)}{(t + 1)(t + h + 1)}] \frac{1}{h}

\lim_{h \rightarrow 0} 4[ \frac{(t + h)(t + 1) - t(t + h + 1)}{(t + h + 1)(t + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} 4[ \frac{(t^2 + ht + t + h - t^2 - ht - t)}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h}

\lim_{h \rightarrow 0} [ \frac{4h}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} ( \frac{4}{t^2 + ht + 2t + h + 1} ) = \frac{4}{t^2 + 2t + 1}

\boxed{G'(t) = \frac{4}{(t + 1)^2}}

Mathematica solution:
- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}
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[Galileo]

\frac{4}{(t+1)^2}=- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}


No mistakes

Check it!

 

[twoflower]

I am attempting to find the derivative of this function using the definition of derivative, however my solution is not the same as Mathematica's solution.
Is my formulation incorrect on some protocol somewhere?
G(t) = \frac{4t}{t + 1}
G'(t) = \lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h}
\lim_{h \rightarrow 0} [(\frac{4(t + h)}{(t + h) + 1}) - (\frac{4t}{t + 1})] \frac{1}{h} = \lim_{h \rightarrow 0} 4 [\frac{(t + h)(t + 1)}{(t + h + 1)(t + 1)} - \frac{t(t + h + 1)}{(t + 1)(t + h + 1)}] \frac{1}{h}
\lim_{h \rightarrow 0} 4[ \frac{(t + h)(t + 1) - t(t + h + 1)}{(t + h + 1)(t + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} 4[ \frac{(t^2 + ht + t + h - t^2 - ht - t)}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h}
\lim_{h \rightarrow 0} [ \frac{4h}{(t^2 + ht + 2t + h + 1)}] \frac{1}{h} = \lim_{h \rightarrow 0} ( \frac{4}{t^2 + ht + 2t + h + 1} ) = \frac{4}{t^2 + 2t + 1}
\boxed{G'(t) = \frac{4}{(t + 1)^2}}
Mathematica solution:
- \frac{4t}{(1 + t)^2} + \frac{4}{t + 1}
[/Color]

It's the same result, so yours is also correct.