1 Crore+ students have signed up on EduRev. Have you? 
Differentiate sin^{2}(θ^{2} + 1) with respect to θ^{2}
y = sin^{2}(θ^{2}+1)
v = θ^{2}
dy/d(v) = dydθ/dvdθ
dy/dthη = sin^{2}(V+1)
= 2sin(V+1)⋅cos(V+1)dv/dθ
= 2sin(θ^{2}+1)cos(θ^{2}+1)
= sin2(θ^{2}+1).
Difference equation in discrete systems is similar to the _____________ in continuous systems.
Difference equation are the equations used in discrete time systems and difference equations are similar to the differential equation in continuous systems.
X = at² and y = 2at are parametric equations of
Together the equations x = at^{2} and y = 2at (where t is the parameter) are called the parametric equations of the parabola y^{2} = 4ax.
If x = 4(t + sin t), y = 4(1cos t), Evaluate dy/dx at t= π/2
x = a(t+sin t)
⟹dx/dt = a(1+cos t)
And y = a(1−cos t)
⟹dy/dt = a[0−(−sin t)]
=a sin t
Therefore, dy/dx = a sin t/(a(1+cos t))
= 2sin t/2 cos t/2)/(2cos^2 t/2)
=tan(t/2)
At π/2
tan(π/4) = 1
Find dy/dx if x = cos^{3} θ, y = sin^{3} θ
x = cos^{3}θ
⇒ dx/dθ = a(3cos^{2}θ)(−sinθ)
= −3sinθcos^{2}θ
y = sin^{3}θ
⇒dy/dθ = a(3sin^{2}θ)(cosθ)
= 3sin^{2}θcosθ
∴dy/dx = (dy/dθ)(dx/dθ)
= (3sin^{2}θcosθ)/(−3asinθcos^{2}θ)
= −sinθ/cosθ
= −tanθ
Find ; x = 20 (cos t + t sin t) and y = 20 ( sin t  t cos t)
x = 20(cost + tsint)
differentiate x with respect to t,
dx/dt = 20{d(cost)/dt + d(tsint)/dt]
= 20[sint + {t. d(sint)/dt + sint.dt/dt}]
= 20[ sint + tcost + sint]
= 20t.cost
hence, dx/dt = 20t.cost (1)
y = 20(sint  tcost)
differentiate y with respect to t,
dy/dt = 20[d(sint)/dt  d(tcost)/dt ]
= 20[cost  {t.d(cost)/dt + cost.dt/dt}]
= 20[cost +tsint cost]
= 20t.sint
hence, dy/dt = 20t.sint (2)
dividing equations (2) by (1),
dy/dx = 20t.sint/20t.cost
dy/dx = tant
now again differentiate with respect to x
d²y/dx² = sec²t. dt/dx (3)
now from equation (1),
dx/dt = 20t.cost
so, dt/dx =1/at.cost put it in equation (3),
e.g., d²y/dx² = sec²t. 1/20t.cost
d²y/dx² = sec³t/20t
If f (x) = [x sin p x] { where [x] denotes greatest integer function}, then f (x) is
If −1≤x≤1, then 0≤xsinπx≤1/2
∴ f(x)=[xsinπx]=0, for −1≤x≤1
If 1<x<1+h, where h is a small positive real number, then
π<πx<π+πh
⇒−1<sinπx<0
⇒−1<xsinπx<0
∴ f(x)=[xsinπx]=−1 in the right neighbourhood of x=1.
Thus, f(x) is constant and equal to zero in [−1,1] and so f(x) is differentiable and hence continuous on (−1,1).
Differentiate sin x^{3} with respect to x^{3}
Differentiation of sinx^{3} with respect to x^{3} is
d(sin(x^{3})/dx = cosx^{3}
209 videos218 docs139 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
209 videos218 docs139 tests









