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dy/dx=d/du (∛(u2 ))×du/dx ,where u=(x2+4)
Q2. If f(x) = x sinx,then find f'(x)
Apply the product rule
d/dx (f(x))g(x)) = d/dx(f(x)) g(x) + f(x)d/dx(g(x)) with f(x) = x and g(x) = sin(x):
f'(x)=d/dx (x sinx)=d/dx (x)*sinx+x*d/dx (sinx)
Q3. Evaluate d/dx ((2 + 3 sinx)/(2 – 3 sinx ))
Apply the quotient rule
d/dx (f(x)/g(x)) = d/dx(f(x) g(x)) – f(x) d/dx(g(x)) / g2(x)
d/dx ((2 + 3 sinx)/(2 – 3 sinx ))=((2-3sinx) d/dx (2+3sinx)-(2+3sinx)d/dx(2-3sinx))/(2-3sinx)2
= 3cosx(2-3sinx+2+3sinx)/(2-3sinx)2 =12cosx/(2-3sinx)2
dx/dθ=d/dθ (θ+sinθ)=1+cosθ dy/dθ=d/dθ (1+cosθ)=-sinθ
By chain rule of differentiation dy/dx=dy/dθ×dθ/dθ
dy/dx=-sinθ×1/(1+cosθ)=-sinθ/(1+cosθ)=-(2 sin(θ/2) cos(θ/2))/(2 cos2(θ/2) )=-tan(θ/2)
Q5. Find dy/dx,where x(2/3)+y(2/3)=a(2/3)
d/dx (x(2/3) )+d/dx (y(2/3) )=d/dx (a(2/3) )
2/3 x(-1/3)+2/3 y(-1/3)*dy/dx=0
dy/dx=(-2/3 x(-1/3))/(2/3 y(-1/3) )=-x(-1/3)/y(-1/3) =-y(1/3)/x(1/3) = -∛(y/x)
Q6. Evaluate ∫(x2+x+1)/(x√x) dx
∫(x2+x+1)/(x√x) dx=∫x2/(x√x) dx+∫x/(x√x) dx+∫1/(x√x) dx
=∫√x dx+∫1/√x dx+∫1/(x√x) dx
=∫x(1/2) dx+∫x(-1/2) dx+∫x(-3/2) dx
= 2/3 x√x+2√x-2/√x+C
Where C is the indefinite integral constant.
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