Page:Calculus Made Easy.pdf/66

 Expand the numerator by the binomial theorem (see p. 141).

$P =\frac{1}{140^5}(40^5 + 5\times 40^4 t + 10 \times 40^3 t^2+10 \times 40^2 t^3+ 5 \times 40t^4+t^5)$;

hence $$\dfrac{dP}{dt} = \dfrac{1}{537,824 \times 10^5}$$, $(5 \times 40^4 + 20 \times 40^3 t + 30 \times 40^2 t^2 + 20 \times 40t^3 + 5t^4)$,

when $$t=100$$ this becomes $$0.036$$ atmosphere per degree Centigrade change of temperature.

Exercises III. (See the Answers on p. 255.)

(1) Differentiate
 * (a) $$u=1+x+\dfrac{x^2}{1 \times 2}+\dfrac{x^3}{1 \times 2 \times 3}+\dotsb$$
 * (b) $$y=ax^2+bx+c$$
 * (c) $$y=(x+a)^2$$
 * (d) $$y=(x+a)^3$$

(2) If $$w=at-\frac{1}{2}bt^2$$, find $$\dfrac{dw}{dt}$$.

(3) Find the differential coefficient of $y=(x+\sqrt{-1})\times (x-\sqrt{-1})$.

(4) Differentiate $y=(197x-34x^2)\times (7+22x-83x^3).$.

(5) If $$x=(y+3) \times (y+5)$$, find $$\dfrac{dx}{dy}$$.

(6) Differentiate $$y = 1.3709x \times (112.6+45.202x^2)$$