Advertisements
Advertisements
प्रश्न
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
Advertisements
उत्तर
y = sin(log x)
y1 = cos(log x) `"d"/"dx"` (log x)
y1 = cos(log x) . `1/x`
∴ xy1 = cos(log x)
Differentiating both sides with respect to x, we get
xy2 + y1(1) = -sin(log x) . \[\frac{1}{x}\]
⇒ x[xy2 + y1] = -sin(log x)
⇒ x2y2 + xy1 = -y
⇒ x2y2 + xy1 + y = 0
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
`sqrtx + 1/root(3)(x) + e^x`
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
`e^x/(1 + x)`
Differentiate the following with respect to x.
`(x^2 + x + 1)/(x^2 - x + 1)`
Find `"dy"/"dx"` for the following function
xy – tan(xy)
Differentiate the following with respect to x.
(sin x)x
Differentiate the following with respect to x.
(sin x)tan x
Find `"dy"/"dx"` of the following function:
x = ct, y = `c/t`
Find `"dy"/"dx"` of the following function:
x = log t, y = sin t
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
