Answer to Question #226680 in Calculus for Portia

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) = "\\int"x√x t√t^2+1dt



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

"1)\\\\\ny=\\frac{1}{\\sqrt{x(x^2-\\frac{2}{x})}}\\\\\ny=\\frac{1}{\\sqrt{x^3-2}}\\\\\n=(x^3-2)^\\frac{-1}{2}\\\\\n\\text{differentiate with respect to x, we get}\\\\\ny'=\\frac{-1}{2}(x^3-2)^\\frac{-3}{2}.3x^2\\space ( \\text{by chain rule })\\\\\n=\\frac{-3x^2}{2(x^3-2)^\\frac{3}{2}}\\\\\n2)\\\\\nh=\\frac{sinx}{1+cosx}\\\\\n\\text{ by using}\\frac{u}{v}\\text{formula and chain rule })\\\\\nh'=\\frac{(1+cosx)cosx-sinx(-sinx)}{(1+cosx)^2}\\\\\n=\\frac{cosx+cos^2x+sin^2x}{(1+cosx)^2}\\\\\n=\\frac{cosx+1}{(1+cosx)^2}\\\\\n=\\frac{1}{1+cosx}\\\\\n\n3)\\\\\ng(x)=(cos 5x)^{sin(x^2)}\\\\\n\\text{taking logarithm both side, we get })\\\\\nlog (g)=sin(x^2) log(cos5x)\\\\\n\\text{differentiate with respect to x, we get}\\\\\n\\frac{g'}{g}=sin(x^2).(\\frac{1}{cos5x}).(-sin5x).(5)+log(cos5x).cos(x^2).(2x)\\space\\text{by chain rule })\\\\\ng'=g[\\frac{-5sin(x^2)sin5x}{cos5x}+2xcos(x^2)log(cos5x)]\\\\\n=(cos 5x)^{sin(x^2)}\n[\\frac{-5sin(x^2)sin5x}{cos5x}+2xcos(x^2)log(cos5x)]\\\\\n\n4)\\\\\nf=\\int_{\\sqrt{x}}^x t\\sqrt{t^2+1}dt\\\\\n\\text{by using leibnitz rule for differentiation, we get}\\\\\nf'=\\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)})\\\\\n=0+1.(x\\sqrt{x^2+1})-\\frac{1}{2\\sqrt{x}}.(\\sqrt{x(x+1)})\\\\\n=x\\sqrt{x^2+1}-\\frac{\\sqrt{x+1}}{2}"


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