Question Number 201764 by hardmath last updated on 11/Dec/23
$$\mathrm{1}.\:\mathrm{y}\:=\:\mathrm{tgx}\:−\:\mathrm{ctgx}\:\:\rightarrow\:\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{2}.\:\mathrm{y}\:=\:\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{arctgx}\:\rightarrow\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{3}.\:\mathrm{y}\:=\:\mathrm{cos}^{\mathrm{4}} \:\mathrm{x}\:\rightarrow\:\mathrm{y}^{'} \:=\:? \\ $$$$\mathrm{4}.\:\begin{cases}{\mathrm{x}\:=\:\mathrm{2t}}\\{\mathrm{y}\:=\:\mathrm{3t}^{\mathrm{2}} \:−\:\mathrm{5t}}\end{cases}\:\:\:\rightarrow\:\:\:\mathrm{x}^{'} \:,\:\mathrm{y}^{'} \:=\:? \\ $$
Answered by Calculusboy last updated on 11/Dec/23
![Solution: (1) y=tgx−ctgx if tgx=tanx and ctgx=cotx y′=sec^2 x+cosec^2 x (2) by using product rule if tgx=tanx y′=[(1+x^2 )(d/dx)arctgx+arctgx(d/dx)(1+^2 )] y′=[(1+x^2 )×(1/(1+x^2 ))+2xarctgx] y′=1+2x arctgx (3)y=cos^4 x let p=cosx y=(cosx)^4 (dp/dx)=−sinx y=p^4 (dy/dp)=4p^3 ⇔ y′=4cos^3 x×−sinx=−4cos^3 xsinx y′=−4cos^3 xsinx](https://www.tinkutara.com/question/Q201773.png)
$$\boldsymbol{{Solution}}:\:\left(\mathrm{1}\right)\:\:\boldsymbol{{y}}=\boldsymbol{{tgx}}−\boldsymbol{{ctgx}}\:\:\boldsymbol{{if}}\:\:\boldsymbol{{tgx}}=\boldsymbol{{tanx}}\:\:\boldsymbol{{and}}\:\boldsymbol{{ctgx}}=\boldsymbol{{cotx}} \\ $$$$\boldsymbol{{y}}'=\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{{cosec}}^{\mathrm{2}} \boldsymbol{{x}} \\ $$$$\left(\mathrm{2}\right)\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{product}}\:\boldsymbol{{rule}}\:\:\boldsymbol{{if}}\:\boldsymbol{{tgx}}=\boldsymbol{{tanx}} \\ $$$$\boldsymbol{{y}}'=\left[\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\boldsymbol{{arctgx}}+\boldsymbol{{arctgx}}\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\left(\mathrm{1}+^{\mathrm{2}} \right)\right] \\ $$$$\boldsymbol{{y}}'=\left[\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)×\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }+\mathrm{2}\boldsymbol{{xarctgx}}\right] \\ $$$$\boldsymbol{{y}}'=\mathrm{1}+\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{arctgx}} \\ $$$$\left(\mathrm{3}\right)\boldsymbol{{y}}=\boldsymbol{{cos}}^{\mathrm{4}} \boldsymbol{{x}}\:\:\boldsymbol{{let}}\:\boldsymbol{{p}}=\boldsymbol{{cosx}} \\ $$$$\boldsymbol{{y}}=\left(\boldsymbol{{cosx}}\right)^{\mathrm{4}} \:\:\:\:\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=−\boldsymbol{{sinx}} \\ $$$$\boldsymbol{{y}}=\boldsymbol{{p}}^{\mathrm{4}} \:\:\:\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dp}}}=\mathrm{4}\boldsymbol{{p}}^{\mathrm{3}} \:\:\Leftrightarrow\:\:\boldsymbol{{y}}'=\mathrm{4}\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}×−\boldsymbol{{sinx}}=−\mathrm{4}\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{xsinx}} \\ $$$$\boldsymbol{{y}}'=−\mathrm{4}\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{xsinx}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Calculusboy last updated on 12/Dec/23
$$\boldsymbol{{you}}\:\:\boldsymbol{{are}}\:\boldsymbol{{welcome}} \\ $$
Commented by hardmath last updated on 12/Dec/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser} \\ $$