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Question Number 74933 by imelder last updated on 04/Dec/19

$$\mathrm{differentiate}\mathrm{the}\mathrm{following}\mathrm{functions}$$$$\mathrm{a})\mathrm{f}\left(\mathrm{x}\right)={2\mathrm{x}}^5\mathrm{coshx}$$

Commented by mathmax by abdo last updated on 04/Dec/19

$$f\left(x\right)=2x^{5}ch\left(x\right)=2x^5\times \frac{e^x+e^{-x}}{2}=x^5(e^x+e^{-x})\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=5x^4(e^x+e^{-x})+x^5(e^x-e^{-x})=(5x^4+x^5)e^x+(5x^4-x^5)e^{-x}$$$$anotherway{f}^{{}^{\prime }}\left(x\right)=10x^{4}ch\left(x\right)+2x^{5}sh\left(x\right)$$

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