Advertisements
Advertisements
प्रश्न
If xm . yn = (x + y)m+n, then show that `"dy"/"dx" = y/x`
Advertisements
उत्तर
xm . yn = (x + y)m+n
Taking logarithm on both sides we get,
m log x + n log y = (m + n) log(x + y)
Differentiating with respect to x,
`m1/x + n1/y "dy"/"dx" = (m + n) 1/(x + y)(1 + "dy"/"dx")`
`"dy"/"dx"(n/y - (m + n)/(x + y)) = (m + n)/(x + y) - m/x`
`"dy"/"dx" ((nx + ny - my - ny)/(y(x + y))) = ((mx + nx - mx - my)/(x(x + y)))`
`"dy"/"dx" ((nx - my)/(y(x + y))) = (nx - my)/(x(x + y))`
`"dy"/"dx" = y/x`
Hence proved.
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.
x3 ex
Differentiate the following with respect to x.
sin2 x
Differentiate the following with respect to x.
`sqrt(1 + x^2)`
Differentiate the following with respect to x.
(ax2 + bx + c)n
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(θ – sin θ), y = a(1 – cos θ)
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
If y = 2 + log x, then show that xy2 + y1 = 0.
If xy . yx , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))`
