Advertisements
Advertisements
Question
Differentiate the following with respect to x.
(sin x)tan x
Advertisements
Solution
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
RELATED QUESTIONS
Differentiate the following with respect to x.
(x2 – 3x + 2) (x + 1)
Differentiate the following with respect to x.
ex (x + log x)
Differentiate the following with respect to x.
sin(x2)
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`
Differentiate the following with respect to x.
(sin x)x
Differentiate the following with respect to x.
`sqrt(((x - 1)(x - 2))/((x - 3)(x^2 + x + 1)))`
Find `"dy"/"dx"` of the following function:
x = a cos3θ, y = a sin3θ
Find y2 for the following function:
y = e3x+2
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`
