Advertisements
Advertisements
प्रश्न
find dy/dx if x=e2t , y=`e^sqrtt`
उत्तर
x=e2t , y=`e^sqrtt`
Differentiating x w.r.t t
`"dx"/"dt"=d(e^(2t))/(dt)=e^(2t)d/"dt"(2t)=2e^(2t)`
Differentiating y w.r.t t
`"dy"/"dt"=d(e^(sqrt t))/(dt)=e^(sqrt t)d/"dt"(sqrt t)=(e^(sqrt t))/(2sqrt t)`
By parametric rule we get
`dy/dx=("dy"/"dt")/("dx"/"dt")`
`=((e^(sqrt t))/(2sqrt t))/(2e^(2t))`
`=(e^sqrtt)/(4sqrt t.e^(2t))`
`e^sqrtt`
APPEARS IN
संबंधित प्रश्न
If `x=a(t-1/t),y=a(t+1/t)`, then show that `dy/dx=x/y`
If x=α sin 2t (1 + cos 2t) and y=β cos 2t (1−cos 2t), show that `dy/dx=β/αtan t`
Find the value of `dy/dx " at " theta =pi/4 if x=ae^theta (sintheta-costheta) and y=ae^theta(sintheta+cos theta)`
Derivatives of tan3θ with respect to sec3θ at θ=π/3 is
(A)` 3/2`
(B) `sqrt3/2`
(C) `1/2`
(D) `-sqrt3/2`
If x and y are connected parametrically by the equation, without eliminating the parameter, find `dy/dx.`
`x = 4t, y = 4/y`
If x and y are connected parametrically by the equation without eliminating the parameter, find `dy/dx`.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
If `x = acos^3t`, `y = asin^3 t`,
Show that `(dy)/(dx) =- (y/x)^(1/3)`
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find `dy/dx` when `theta = pi/3`
Evaluate : `int (sec^2 x)/(tan^2 x + 4)` dx
x = `"t" + 1/"t"`, y = `"t" - 1/"t"`
sin x = `(2"t")/(1 + "t"^2)`, tan y = `(2"t")/(1 - "t"^2)`
x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`
If x = ecos2t and y = esin2t, prove that `"dy"/"dx" = (-y log x)/(xlogy)`
If x = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that `("dy"/"dx")_("at t" = pi/4) = "b"/"a"`
Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0
If x = sint and y = sin pt, prove that `(1 - x^2) ("d"^2"y")/("dx"^2) - x "dy"/"dx" + "p"^2y` = 0
If x = `a[cosθ + logtan θ/2]`, y = asinθ then `(dy)/(dx)` = ______.
