Advertisements
Advertisements
प्रश्न
Find y2 for the following function:
y = log x + ax
Advertisements
उत्तर
y = log x + ax
`y_1 = "dy"/"dx" = 1/x + "a"^x log "a"` ....`[because "d"/"dx" ("a"^x) = "a"^x log "a"]`
`y_2 = ("d"^2y)/"dx"^2 = "d"/"dx"(1/x) + log "a" "d"/"dx" ("a"^x)`
`= (-1)/x^2 + (log "a")("a"^x log "a")`
`= (-1)/x^2 + ("a"^x log "a")^2`
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
3x4 – 2x3 + x + 8
Differentiate the following with respect to x.
`(3 + 2x - x^2)/x`
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
`(x^2 + x + 1)/(x^2 - x + 1)`
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
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.
xsin x
Differentiate sin3x with respect to cos3x.
