Advertisements
Advertisements
प्रश्न
If x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ), "then show that" "dy"/"dx" = -y/x`.
Advertisements
उत्तर
x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ)`
∴ `x/a = sqrt(secθ - tanθ), y/a = sqrt(secθ + tanθ)`
∴ `sec θ - tanθ = x^2/a^2` ...(1)
`sec θ + tanθ = y^2/a^2` ...(2)
Adding (1) and (2), we get
2secθ = `x^2/a^2 + y^2/a^2`
= `(x^2 + y^2)/a^2`
∴ secθ = `(x^2 + y^2)/(2a^2)`
Subtracting (1) from (2), we get
2tanθ = `y^2/a^2 - x^2/a^2`
= `(y^2 - x^2)/a^2`
∴ tanθ = `(y^2 - x^2)/(2a^2)`
∴ sec2θ - tan2θ = 1 gives,
`((x^2 + y^2)/(2a^2))^2 - ((y^2 - x^2)/(2a^2))^2` = 1
∴ (x2 + y2)2 - (y2 - x2)2 = 4a4
∴ (x4 + 2x2y2 + y4) - (y4 - 2x2y2 + x4) = 4a4
∴ 4x2y2 = 4a4
∴ x2y2 = a4
Differentiating both sides w.r.t. x, we get
`x^2."d"/"dx"(y^2) + y^2."d"/"dx"(x^2)` = 0
∴ `x^2 xx 2y"dy"/"dx" + y^2 xx 2x` = 0
∴ `2x^2y"dy"/"dx"` = -2xy2
∴ `"dy"/"dx" = -y/x`.
APPEARS IN
संबंधित प्रश्न
Differentiate the following function with respect to x: `(log x)^x+x^(logx)`
Differentiate the function with respect to x.
`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`
Differentiate the function with respect to x.
`(x + 1/x)^x + x^((1+1/x))`
Differentiate the function with respect to x.
`(sin x)^x + sin^(-1) sqrtx`
Differentiate the function with respect to x.
`(x cos x)^x + (x sin x)^(1/x)`
Find `bb(dy/dx)` for the given function:
xy + yx = 1
Find `bb(dy/dx)` for the given function:
xy = `e^((x - y))`
Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
- By using the product rule.
- By expanding the product to obtain a single polynomial.
- By logarithmic differentiation.
Do they all give the same answer?
Differentiate the function with respect to x:
xx + xa + ax + aa, for some fixed a > 0 and x > 0
Evaluate
`int 1/(16 - 9x^2) dx`
Find `dy/dx` if y = xx + 5x
Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`
xy = ex-y, then show that `"dy"/"dx" = ("log x")/("1 + log x")^2`
If y = (log x)x + xlog x, find `"dy"/"dx".`
If `log_10((x^3 - y^3)/(x^3 + y^3))` = 2, show that `dy/dx = -(99x^2)/(101y^2)`.
`"If" y = sqrt(logx + sqrt(log x + sqrt(log x + ... ∞))), "then show that" dy/dx = (1)/(x(2y - 1).`
If y = `x^(x^(x^(.^(.^.∞))`, then show that `"dy"/"dx" = y^2/(x(1 - logy).`.
If x = esin3t, y = ecos3t, then show that `dy/dx = -(ylogx)/(xlogy)`.
If x = a cos3t, y = a sin3t, show that `"dy"/"dx" = -(y/x)^(1/3)`.
If x = 2cos4(t + 3), y = 3sin4(t + 3), show that `"dy"/"dx" = -sqrt((3y)/(2x)`.
If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.
Find the nth derivative of the following: log (ax + b)
Choose the correct option from the given alternatives :
If xy = yx, then `"dy"/"dx"` = ..........
If y = A cos (log x) + B sin (log x), show that x2y2 + xy1 + y = 0.
If f(x) = logx (log x) then f'(e) is ______
If y = `25^(log_5sin_x) + 16^(log_4cos_x)` then `("d"y)/("d"x)` = ______.
If y = log [cos(x5)] then find `("d"y)/("d"x)`
If y = 5x. x5. xx. 55 , find `("d"y)/("d"x)`
Derivative of loge2 (logx) with respect to x is _______.
If xy = ex-y, then `"dy"/"dx"` at x = 1 is ______.
`d/dx(x^{sinx})` = ______
`"d"/"dx" [(cos x)^(log x)]` = ______.
Derivative of `log_6`x with respect 6x to is ______
`8^x/x^8`
`lim_("x" -> 0)(1 - "cos x")/"x"^2` is equal to ____________.
`lim_("x" -> -2) sqrt ("x"^2 + 5 - 3)/("x" + 2)` is equal to ____________.
If `"y" = "e"^(1/2log (1 + "tan"^2"x")), "then" "dy"/"dx"` is equal to ____________.
If `f(x) = log [e^x ((3 - x)/(3 + x))^(1/3)]`, then `f^'(1)` is equal to
If y = `(1 + 1/x)^x` then `(2sqrt(y_2(2) + 1/8))/((log 3/2 - 1/3))` is equal to ______.
If y = `x^(x^2)`, then `dy/dx` is equal to ______.
Derivative of log (sec θ + tan θ) with respect to sec θ at θ = `π/4` is ______.
The derivative of x2x w.r.t. x is ______.
Evaluate:
`int log x dx`
