Advertisements
Advertisements
प्रश्न
Find y2 for the following function:
y = log x + ax
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.
x3 ex
Differentiate the following with respect to x.
`e^x/(1 + x)`
Differentiate the following with respect to x.
ex sin x
Differentiate the following with respect to x.
sin(x2)
If 4x + 3y = log(4x – 3y), then find `"dy"/"dx"`
Differentiate the following with respect to x.
xsin x
Find y2 for the following function:
x = a cosθ, y = a sinθ
If y = 2 + log x, then show that xy2 + y1 = 0.
If y = `(x + sqrt(1 + x^2))^m`, then show that (1 + x2) y2 + xy1 – m2y = 0
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
