Question #226680

Find the derivative of the following functions by using the appropriate rultof differentiation:

1. y=1/√x[x^2-2/X]

2. h(X)=sin x/1+cos x

3. G(X)=(cos5x)^sin(x^2)

4. F(X) = \intx√x t√t^2+1dt



1
Expert's answer
2021-08-30T17:27:00-0400

1)y=1x(x22x)y=1x32=(x32)12differentiate with respect to x, we gety=12(x32)32.3x2 (by chain rule )=3x22(x32)322)h=sinx1+cosx by usinguvformula and chain rule )h=(1+cosx)cosxsinx(sinx)(1+cosx)2=cosx+cos2x+sin2x(1+cosx)2=cosx+1(1+cosx)2=11+cosx3)g(x)=(cos5x)sin(x2)taking logarithm both side, we get )log(g)=sin(x2)log(cos5x)differentiate with respect to x, we getgg=sin(x2).(1cos5x).(sin5x).(5)+log(cos5x).cos(x2).(2x) by chain rule )g=g[5sin(x2)sin5xcos5x+2xcos(x2)log(cos5x)]=(cos5x)sin(x2)[5sin(x2)sin5xcos5x+2xcos(x2)log(cos5x)]4)f=xxtt2+1dtby using leibnitz rule for differentiation, we getf=xxx(tt2+1)dt+dxdx.(xx2+1)dxdx.(x(x+1))=0+1.(xx2+1)12x.(x(x+1))=xx2+1x+121)\\ y=\frac{1}{\sqrt{x(x^2-\frac{2}{x})}}\\ y=\frac{1}{\sqrt{x^3-2}}\\ =(x^3-2)^\frac{-1}{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}}\\ 2)\\ 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}\\ 3)\\ 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)]\\ 4)\\ 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}


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