Answer to Question #228462 in Calculus for Neliswa

Question

Find the derivatives of the following functions by using the appropriate rules of differentiation:

(i) y= 1/√x[x^2-2/x]

(ii) g(x)=(cos 5x)^sin(x^2)

(iii) h (x)=sinx/1+cos x

(iv) f(x)= $\int$ _√x^xt√t^2+1dt

Answer 1

$i) y=\frac{1}{\sqrt{x(x^2-\frac{2}{x})}}\\ y=\frac{1}{\sqrt{x^3-2}}\\ \text{differentiate with respect to x, we get}\\ y'=\frac{-1}{2}(x^3-2)^\frac{-3}{2}.3x^2\space ( \text{by chain rule })\\ =\frac{-3x^2}{2(x^3-2)^\frac{3}{2}}\\ ii)\\ g(x)=(cos 5x)^{sin(x^2)}\\ \text{taking logarithm both side, we get })\\ log (g)=sin(x^2) log(cos5x)\\ \text{differentiate with respect to x, we get}\\ \frac{g'}{g}=sin(x^2).(\frac{1}{cos5x}).(-sin5x).(5)+log(cos5x).cos(x^2).(2x)\space\text{by chain rule })\\ g'=g[\frac{-5sin(x^2)sin5x}{cos5x}+2xcos(x^2)log(cos5x)]\\ =(cos 5x)^{sin(x^2)} [\frac{-5sin(x^2)sin5x}{cos5x}+2xcos(x^2)log(cos5x)]\\ iii)\\ h=\frac{sinx}{1+cosx}\\ \text{ by using}\frac{u}{v}\text{formula and chain rule })\\ h'=\frac{(1+cosx)cosx-sinx(-sinx)}{(1+cosx)^2}\\ =\frac{cosx+cos^2x+sin^2x}{(1+cosx)^2}\\ =\frac{cosx+1}{(1+cosx)^2}\\ =\frac{1}{1+cosx}\\ iv)\\ f=\int_{\sqrt{x}}^x t\sqrt{t^2+1}dt\\ \text{by using leibnitz rule for differentiation, we get}\\ f'=\int_{\sqrt{x}}^x \frac{\partial}{\partial x}(t\sqrt{t^2+1})dt+\frac{dx}{dx}.(x\sqrt{x^2+1})-\frac{d\sqrt{x}}{dx}.(\sqrt{x(x+1)})\\ =0+1.(x\sqrt{x^2+1})-\frac{1}{2\sqrt{x}}.(\sqrt{x(x+1)})\\ =x\sqrt{x^2+1}-\frac{\sqrt{x+1}}{2}$