Advertisements
Advertisements
प्रश्न
Differentiate the following with respect to x.
(sin x)x
Advertisements
उत्तर
Let y = (sin x)x
Taking logarithm on both sides we get,
log y = x log(sin x)
Differentiating with respect to x,
`1/y * "dy"/"dx" = x "d"/"dx" log(sin x) + log(sin x) "d"/"dx" (x)`
`1/y * "dy"/"dx" = x1/(sin x) (cos x) + log(sin x)(1)`
`1/y * "dy"/"dx"` = x cot x + log(sin x)
`"dy"/"dx"` = y[x cot x + log(sin x)]
`"dy"/"dx"`= (sin x)x [x cot x + log(sin x)]
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
`5/x^4 - 2/x^3 + 5/x`
Differentiate the following with respect to x.
`(3 + 2x - x^2)/x`
Differentiate the following with respect to x.
`(sqrtx + 1/sqrtx)^2`
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Find `"dy"/"dx"` for the following function
xy – tan(xy)
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Differentiate the following with respect to x.
(sin x)tan x
Find `"dy"/"dx"` of the following function:
x = a(θ – sin θ), y = a(1 – cos θ)
Find y2 for the following function:
y = e3x+2
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
