Advertisements
Advertisements
प्रश्न
Differentiate the following with respect to x.
(sin x)tan x
Advertisements
उत्तर
Let y = (sin x)tan x
Taking logarithm on both sides we get,
log y = tan x log(sin x)
Differentiating with respect to x,
`1/y "dy"/"dx" = tan x "d"/"dx"` (log (sin x)) + log (sin x) `"d"/"dx"` (tan x)
= tan x `1/(sin x)`(cos x) + log (sin x) sec2x
`1/y "dy"/"dx" = (sin x)/(cos x) xx (cos x)/(sin x)` + log (sin x)(sec2x)
= 1 + log (sin x)(sec2x)
`"dy"/"dx"` = y[1 + sec2x log (sin x)]
= (sin x)tan x[1 + sec2x 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.
`sqrtx + 1/root(3)(x) + e^x`
Differentiate the following with respect to x.
x4 – 3 sin x + cos x
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
`1/sqrt(1 + x^2)`
If `xsqrt(1 + y) + ysqrt(1 + x)` = 0 and x ≠ y, then prove that `"dy"/"dx" = - 1/(x + 1)^2`
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Find `"dy"/"dx"` of the following function:
x = a cos3θ, y = a sin3θ
Find y2 for the following function:
y = e3x+2
Find y2 for the following function:
y = log x + ax
