Question
Differentiate the following from the first principle
$\sin(2\text{x}-3)$
$\sin(2\text{x}-3)$
$\text{f}(\text{x})=\sin(\text{2x}-3)$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\big\{2(\text{x}+\text{h})-3\big\}-\sin(\text{2x}-3)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\cos\frac{(\text{2x}+\text{2h}-3+\text{2x}-3)}{2}\times\sin\frac{(\text{2x}+\text{2h}-3-2-\text{x}+3)}{2}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}2\cos(\text{2x}-3+\text{h}).\frac{\sin\text{x}}{\text{x}}$
$=2\cos(\text{2x}-3)$
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