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san4es73 [151]
3 years ago
11

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)
Mathematics
1 answer:
nadya68 [22]3 years ago
3 0

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)

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