Question

Find the derivative of the given function with respect to the independent variable x or t. The symbols a, b and c are constants greater than 1. You are not required to combine like terms, reduce fractions, or otherwise simplify your final answer.
(1) y = (2/[ a+bx])^3
(2) y = (at^3 - 3bt)^3
(3) y = (t^b)e^(b/t)
(4) z = ax^2.sin (4x)

Answer 1

Answer:

(1) $y=(\frac{2}{a+bx})^3$

By differentiating w.r.t. x,

$\frac{dy}{dx}=3(\frac{2}{a+bx})^2\times \frac{d}{dt}(\frac{2}{a+bx})$

$=3(\frac{2}{a+bx})^2\times (-\frac{2}{(a+bx)^2})$

$=-\frac{24}{(a+bx)^4}$

(2) $y=(at^3-3bt)^3$

By differentiating w.r.t. t,

$\frac{dy}{dt}=3(at^3-3bt)^2\times \frac{d}{dt}(at^3-3bt)$

$=3(at^3-3bt)^2 (3at^2-3b)$

$=9t^2(at^2-3b)^2(at^2-b)$

(3) $y=(t^b)(e^\frac{b}{t})$

Differentiating w.r.t. t,

$\frac{dy}{dt}=t^b\times \frac{d}{dt}(e^\frac{b}{t})+\frac{d}{dt}(t^b)\times e^\frac{b}{t}$

$=t^b(e^\frac{b}{t})\times \frac{d}{dt}(\frac{b}{t}) + bt^{b-1}(e^\frac{b}{t})$

$=t^be^\frac{b}{t}(-\frac{b}{t^2})+bt^{b-1}e^{\frac{b}{t}}$

(4) $z = ax^2.sin (4x)$

Differentiating w.r.t. x,

$\frac{dz}{dt}=ax^2\times \frac{d}{dx}(sin (4x))+sin (4x)\times \frac{d}{dx}(ax^2)$

$=ax^2\times cos(4x).4+sin (4x)(2ax)$

$=4ax^2cos (4x)+2ax sin (4x)$