Advertisements
Advertisements
प्रश्न
Find `"dy"/"dx"` of the following function:
x = log t, y = sin t
उत्तर
x = log t y = sin t
`"dx"/"dt" = 1/"t"` `"dy"/"dt"` = cos t
`"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
`= (cos "t")/(1/"t")` = t cos t
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
(x2 – 3x + 2) (x + 1)
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Differentiate the following with respect to x.
(ax2 + bx + c)n
Find `"dy"/"dx"` for the following function.
x2 – xy + y2 = 1
Differentiate the following with respect to x.
`sqrt(((x - 1)(x - 2))/((x - 3)(x^2 + x + 1)))`
Find y2 for the following function:
y = e3x+2
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`