Respuesta :
Answer:
(1) [tex]y=(\frac{2}{a+bx})^3[/tex]
By differentiating w.r.t. x,
[tex]\frac{dy}{dx}=3(\frac{2}{a+bx})^2\times \frac{d}{dt}(\frac{2}{a+bx})[/tex]
[tex]=3(\frac{2}{a+bx})^2\times (-\frac{2}{(a+bx)^2})[/tex]
[tex]=-\frac{24}{(a+bx)^4}[/tex]
(2) [tex]y=(at^3-3bt)^3[/tex]
By differentiating w.r.t. t,
[tex]\frac{dy}{dt}=3(at^3-3bt)^2\times \frac{d}{dt}(at^3-3bt)[/tex]
[tex]=3(at^3-3bt)^2 (3at^2-3b)[/tex]
[tex]=9t^2(at^2-3b)^2(at^2-b)[/tex]
(3) [tex]y=(t^b)(e^\frac{b}{t})[/tex]
Differentiating w.r.t. t,
[tex]\frac{dy}{dt}=t^b\times \frac{d}{dt}(e^\frac{b}{t})+\frac{d}{dt}(t^b)\times e^\frac{b}{t}[/tex]
[tex]=t^b(e^\frac{b}{t})\times \frac{d}{dt}(\frac{b}{t}) + bt^{b-1}(e^\frac{b}{t})[/tex]
[tex]=t^be^\frac{b}{t}(-\frac{b}{t^2})+bt^{b-1}e^{\frac{b}{t}}[/tex]
(4) [tex]z = ax^2.sin (4x)[/tex]
Differentiating w.r.t. x,
[tex]\frac{dz}{dt}=ax^2\times \frac{d}{dx}(sin (4x))+sin (4x)\times \frac{d}{dx}(ax^2)[/tex]
[tex]=ax^2\times cos(4x).4+sin (4x)(2ax)[/tex]
[tex]=4ax^2cos (4x)+2ax sin (4x)[/tex]
