Advertisements
Advertisements
प्रश्न
Find the derivatives of the following:
x = `"a" cos^3"t"` ; y = `"a" sin^3"t"`
Advertisements
उत्तर
x = a cost , y = a sin3t
`("d"x)/("d"t) = 3"a"cos^2"t" (- sin"t")`, `("d"y)/("dt") = 3 "a"sin^2"t" (cos "t")`
`(("d"y)/("dt"))/(("d"x)/("dt")) = (3"a" sin^2"t" cos"t")/(- 3"a" cos^2"t" sin"t")`
`("d"y)/("d"x) = - (sin "t")/(cos "t")`
`("d"y)/("d"x)` = – tan t
APPEARS IN
संबंधित प्रश्न
Find the derivatives of the following functions with respect to corresponding independent variables:
y = cos x – 2 tan x
Find the derivatives of the following functions with respect to corresponding independent variables:
g(t) = t3 cos t
Find the derivatives of the following functions with respect to corresponding independent variables:
g(t) = 4 sec t + tan t
Find the derivatives of the following functions with respect to corresponding independent variables:
y = (x2 + 5) log(1 + x) e–3x
Differentiate the following:
y = `root(3)(1 + x^3)`
Differentiate the following:
y = sin (ex)
Differentiate the following:
y = (2x – 5)4 (8x2 – 5)–3
Differentiate the following:
s(t) = `root(4)(("t"^3 + 1)/("t"^3 - 1)`
Differentiate the following:
y = `5^((-1)/x)`
Find the derivatives of the following:
`sqrt(x) = "e"^((x - y))`
Find the derivatives of the following:
xy = yx
Find the derivatives of the following:
(cos x)log x
Find the derivatives of the following:
x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)`
Find the derivatives of the following:
sin-1 (3x – 4x3)
Find the derivatives of the following:
`tan^-1 ((cos x + sin x)/(cos x - sin x))`
Find the derivatives of the following:
Find the derivative with `tan^-1 ((sinx)/(1 + cos x))` with respect to `tan^-1 ((cosx)/(1 + sinx))`
Find the derivatives of the following:
If x = a(θ + sin θ), y = a(1 – cos θ) then prove that at θ = `pi/2`, yn = `1/"a"`
Choose the correct alternative:
If y = `1/("a" - z)`, then `("d"z)/("d"y)` is
Choose the correct alternative:
If y = cos (sin x2), then `("d"y)/("d"x)` at x = `sqrt(pi/2)` is
Choose the correct alternative:
x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)` then `("d"y)/("d"x)` is
