Question
Differentiate the functions with respect to 'x'.
$\cos(\text{x}^{2}+1)$

Answer

Let $\text{f}(\text{x})=\cos(\text{x}^{2}+1)\ ...(\text{i})$
$\Rightarrow \text{f}(\text{x}\Delta\text{x})=\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big]\ ...(\text{ii})$ 
Subtracting eq. (i) from eq. (ii) we get
$\text{f}(\text{x}+\Delta\text{x}-\text{f}(\text{x})=\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big]-\cos(\text{x}^{2}+1)$
Dividing both sides by
$\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}=\frac{\cos\big[(\text{x}+\Delta\text{x}^{2})+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\cos\big[(\text{x}+\Delta\text{x}^{2}+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$\text{f}'(\text{x})=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\cos\big[(\text{x}+\Delta\text{x}^{2}+1\big] -\cos(\text{x}^{2}+1)} {\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\frac{(\text{x}+\Delta\text{x})^{2}+1-\text{x}^{2}-1}{2}\Big]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\frac{\text{x}^{2}+\Delta\text{x}^{2}+2\text{x}-\Delta\text{x}-\text{x}^{2}}{2}\Big]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{-2\sin\bigg[\text{x}^{2}+\frac{\Delta\text{x}^{2}}{2}+\text{x}\Delta\text{x}+1\bigg]\sin\bigg[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\bigg]}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\sin\Big[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\Big]}{\Delta\big[\frac{\Delta\text{x}+2\text{x}}{2}\big]}\times\Big(\frac{\Delta\text{x}+2\text{x}}{2}\Big)$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}-2\sin\bigg[\text{x}^{2}+\frac{\Delta\text{x}^{2}}{2}+\text{x}\Delta\text{x}+1\bigg]\cdot\frac{\sin\Big[\Delta\text{x}\frac{(\Delta\text{x}+2\text{x})}{2}\Big]}{\Delta\text{x}\Big[\frac{\Delta\text{x}+2\text{x}}{2}\Big]}\times\bigg[\frac{\Delta\text{x}+2\text{x}}{2}\bigg] $
Taking limit, we have
$=-2\sin(\text{x}^{2}+1)\cdot1\cdot(\text{x})$
$=-2\sin(\text{x}^{2}+1)$
Hence, the required answer is $-2\sin(\text{x}^{2}+1).$

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