Menu Close

Question-117960




Question Number 117960 by andilizhaa last updated on 14/Oct/20
Answered by mathmax by abdo last updated on 14/Oct/20
if you want derivative we have  f(x)=5x^4  +x(√x)+(6/( (√x))) −sinx−2cosx ⇒  f(x) =5x^4  +x^(3/2)  +6 x^(−(1/2)) −sinx −2cosx ⇒  f^′ (x)=20 x^3  +(3/2)x^((3/2)−1)  −3 x^(−(3/2)) −cosx +2sinx  =20 x^3  +(3/2)(√x)−(3/(x(√x))) −cosx+2sinx
$$\mathrm{if}\:\mathrm{you}\:\mathrm{want}\:\mathrm{derivative}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{5x}^{\mathrm{4}} \:+\mathrm{x}\sqrt{\mathrm{x}}+\frac{\mathrm{6}}{\:\sqrt{\mathrm{x}}}\:−\mathrm{sinx}−\mathrm{2cosx}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{5x}^{\mathrm{4}} \:+\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:+\mathrm{6}\:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{sinx}\:−\mathrm{2cosx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\mathrm{20}\:\mathrm{x}^{\mathrm{3}} \:+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} \:−\mathrm{3}\:\mathrm{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{cosx}\:+\mathrm{2sinx} \\ $$$$=\mathrm{20}\:\mathrm{x}^{\mathrm{3}} \:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{x}}−\frac{\mathrm{3}}{\mathrm{x}\sqrt{\mathrm{x}}}\:−\mathrm{cosx}+\mathrm{2sinx} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *