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251.

Differentiate w.r.t `x: sin^(-1) ((2^(x+1))/(1+4^x))`

Answer» Correct Answer - `(2^(x+1)(log2))/((1+4^(x)))`
Let `y=sin^(-1){(2^(x).2)/(1+(2^(x))^(2))}`. Putting `2^(x)=tan theta`, weget
`y=sin^(-1){(2tan theta)/(1+tan^(2)theta)}=sin^(-1)(sin 2theta)=2theta=2tan^(-1)2^(x).`
`therefore(dy)/(dx)=2.(d)/(dx)(tan^(-1)2^(x))=2.(1)/({1+(2^(x))^(2)}).(d)/(dx)(2^(x))=(2)/((1+4^(x))).2^(x)(log2)=(2^(x+1)(log2))/((1+4^(x)))`
Discussion
252.

Find `(dy)/(dx)`, when: `(x^(2)+y^(2))^(2)=xy`

Answer» Correct Answer - `((y-4xy^(2)-4x^(3)))/((4y^(3)+4x^(2)y-x))`
`x^(4)+y^(4)+2x^(2)y^(2)=xy`
`rArr4x^(3)+4y^(3).(dy)/(dx)+4x^(2)y.(dy)/(dx)+4xy^(2)=x(dy)/(dx)+y`
`rArr(4y^(3)+4x^(2)y-x)(dy)/(dx)=(y-4xy^(2)-4x^(3)).`
Discussion
253.

Find `(dy)/(dx)`, when: `x^(2)+y^(2)=log(xy)`

Answer» Correct Answer - `(y(1-2x^(2)))/(x(2y^(2)-1))`
`x^(2)+y^(2)=logx+logy`
`rArr2x+2y(dy)/(dx)=(1)/(x)+(1)/(y).(dy)/(dx).`
Discussion
254.

Differentiate with respect to x:tan3 x

Answer»

Chain rule:

If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx

Let y = tan3 x

dy/dx = 3 tan2 x [d/dx (tan x)]

= 3 tan2 x sec2x
Discussion
255.

Differentiate with respect to x.tan √x

Answer»

Chain rule:

If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx

Let y = tan √x

dy/dx = sec2√x d/dx (√x)

= sec2√x {1/(2√x) }

= sec2√x / 2√x
Discussion
256.

If `(d)/(dx)(log_(e)x)=(1)/(x)` then ` (d)/(dx)(log_(10)x)=`A. `(1)/(x)log_(10)e`B. `(1)/(10)log_(10)e`C. `(1)/(x)log_(e)10`D. `(1)/(e)log_(10)x`

Answer» Correct Answer - A
Discussion
257.

If `n lt 1 ` then `(dx^(n))/(dx)` exists when -A. `x = - 1 `B. ` x = 1`C. ` x = 0 `D. ` x ne 0 `

Answer» Correct Answer - D
Discussion
258.

If `y = cos ^(2(x)/(2))` , state which of the following is the value of `(dy)/(dx)` ?A. cos xB. `(1)/(2) cos x `C. `-(1)/(2) sin x `D. `- sin x `

Answer» Correct Answer - C
Discussion
259.

If `y = cos ^(2(x)/(2))` , state which of the following is the value of `(dy)/(dx)` ?A. sin xB. `(1)/(2) sin x `C. cos xD. `(1)/(2) cos x `

Answer» Correct Answer - B
Discussion
260.

If `y = f(x) = 2 ` then the increment of the function `= Delta y = `A. 1B. `-1`C. 0D. 2

Answer» Correct Answer - C
Discussion
261.

`cos 5x`

Answer» Correct Answer - `-5 sin 5x`
Discussion
262.

If `y= cos^(-1)((2x)/(1+x^(2)))`, then `(dy)/(dx)` is -A. `(-2)/(1+x^(2))` for all xB. `(-2)/(1+x^(2))` for all `|x| lt 1`C. `(2)/(1+x^(2))` for `|x| gt 1`D. none of these

Answer» Correct Answer - B, C
Discussion
263.

If `Delta y ` be the increment of the function `y = f (x)` corresponding to the increment `Delta x ` of x then `Delta y = `A. `f(x-Deltax)-f(x)`B. `f(x+Deltax)-f(x)`C. `f(x+Deltax)- f(Deltax)`D. Nonr of these

Answer» Correct Answer - B
Discussion
264.

`sqrt(sinx^(3))`

Answer» Correct Answer - `(3)/(2)x^(2).(cosx^(3))/(sqrt(sinx))`
Discussion
265.

`tan3x`

Answer» Correct Answer - `3sec^(2)3x`
Discussion
266.

If `f(x)=sin^(-1) (cos x)` then answer the following. Find the value of `f(10)`.A. `10-(7pi)/(2)`B. `(7pi)/(5)-11`C. `(5pi)/(2)-11`D. none of these

Answer» Correct Answer - A
Discussion
267.

`tan^(3)x`

Answer» Correct Answer - `3 tan^(2)xsec^(2)x`
Discussion
268.

`sqrt(x sinx)`

Answer» Correct Answer - `((xcos x+sinx))/(2sqrt(x sinx))`
Discussion
269.

`cos x^(3)`

Answer» Correct Answer - `-3x^(2)sinx^(3)`
Discussion
270.

Let `y=x^(3)-8x+7` and `x=f(t)` given t=0, x=3 Find the value of `(dx)/(dt)`A. `(2)/(19)`B. `(-2)/(19)`C. `(19)/(2)`D. `(-19)/(2)`

Answer» Correct Answer - A
Discussion
271.

Let `y=x^(3)-8x+7` and `x=f(t)` given t=0, x=3 Find the value of `(dy)/(dt)`.A. -2B. 2C. `(1)/(2)`D. `-(1)/(2)`

Answer» Correct Answer - B
Discussion
272.

Let `y=x^(3)-8x+7` and `x=f(t)` given t=0, x=3 Find the value of `(dy)/(dx)`A. -19B. `(1)/(19)`C. 19D. `-(1)/(19)`

Answer» Correct Answer - C
Discussion
273.

`sqrt(cotsqrtx)`

Answer» Correct Answer - `-("cosec"^(2)sqrtx)/(4sqrtxsqrt(cotsqrtx))`
Discussion
274.

Differentiate `(xcosx)^x`with respect to `xdot`

Answer» Correct Answer - `(x cos x)^(x)[( log x +1) +{ log cos x- x cot x}]`
Let `y = (x cos x)^(x).`
Taking logarithm on both sides, we get
`log y = log (x cos x)^(x)`
`=x log ( x cos x )`
`=x log x+ x log cos x`
Differentiating both sides with respect to x, we get
`(1)/(y)(dy)/(dx)=xcdot(1)/(x)+ log x+log cos x+xcdot ((-sin x)/(cos x))`
`" or "(dy)/(dx)=( x cos x)^(x)[(log x +1)+{ log cos x - x cot x}]`
Discussion
275.

`"Find "(dy)/(dx)" for "y=sin(x^(2)+1).`

Answer» `"Let "y= sin(x^(2)+1).`
Putting `u=x^(2)+1," we get y sin u"`
`therefore" "(dy)/(dx)=cos u and (du)/(dx) = 2x`
`"Now, "(du)/(dx)=(dy)/(du).(du)/(dx)`
`= (cos u) (2x)=2x cos(x^(2)+1)`
Discussion
276.

If the equation `2x^(4)+x^(3)-2x-1=0` has two real roots `alpha` and `beta`, then between `alpha` and `beta` the equation `8x^(3)+3x^(2)-2=0` has -A. no rootB. at least one rootC. exactly one rootD. atmost two roots

Answer» Correct Answer - B
Discussion
277.

`cot^(3)x^(2)`

Answer» Correct Answer - `-6x cot^(2)x^(2)"cosec"^(2)x^(2)`
Discussion
278.

Find `(dy)/(dx)` if `x=cos theta - cos 2 theta` `and" "y = sin theta - sin 2theta`

Answer» Correct Answer - `(cos theta -2 cos 2 theta)/(2 sin 2 theta - sin theta)`
The given equations are `x=cos theta - cos 2 theta`
`and" "y = sin theta - sin 2theta`
`"Then, "(dx)/(d""theta)=-sin theta - (-2 sin 2theta)=2 sin 2 theta- sin theta`
`"And "(dy)/(d""theta)=cos theta -2 cos 2theta`
`therefore" "(dy)/(dx)=(dy//d""theta)/(dx//d""theta)=(cos theta-2 cos 2theta)/(2 sin 2theta- sin theta)`
Discussion
279.

If `y=(1)/(3x-1)`, then the value of `[(dy)/(dx)]_(x=0)` is -A. -3B. `(1)/(3)`C. `-(1)/(9)`D. 3

Answer» Correct Answer - A
Discussion
280.

`ax^2+2hxy+by^2+2gx+2fy+c=0`

Answer» Given: `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0." …(i)"`
Differentiating both sides of (i) w.r.t. x, we get
`2ax+2h(x.(dy)/(dx)+y.1)+2by.(dy)/(dx)+2g+2f.(dy)/(dx)=0`
`rArr(2ax+2hy+2g)+(2hx+2by+2f).(dy)/(dx)=0`
`rArr (dy)/(dx)=-((ax+hy+g)/(hx+by+f)).`
Discussion
281.

if `2^x+2^y=2^(x+y)` then the value of `(dy)/(dx)` at `x=y=1`

Answer» Correct Answer - B
Discussion
282.

If `y=10^(10^(x))`, then the value of `(dy)/(dx)` is -A. `10^(10^(x))*(log_(e)10)^(2)`B. `10^(10^(x))*log_(e)10`C. `10^(10^(x))*10^(x)*log_(e)10`D. `10^(10^(x))*10^(x)*(log_(e)10)^(2)`

Answer» Correct Answer - D
Discussion
283.

If `x=e^y+e^((y+ tooo))`, where `x >0,t h e nfin d(dy)/(dx)`A. `(1+x)/(x)`B. `(1)/(x)`C. `(1-x)/(x)`D. `(x)/(1+x)`

Answer» Correct Answer - C
We have,
`x=e^(y+e^(y+e^(y+..." to "oo)))`
`implies" "x=e^(y+x)`
`implies" "log_(e)x=(y+x)`
`implies" "(1)/(x)=(dy)/(dx)+1" "["Differentiating with respect to x"]`
`implies" "(dy)/(dx)=(1-x)/(x)`
Discussion
284.

If `y=sin^(-1)(x/2)+cos^(-1)(x/2)` then `(dy)/(dx)=`A. 1B. -1C. 0D. 2

Answer» Correct Answer - C
Discussion
285.

If `y=x+e^x ,`then `(d^2x)/(dy^2)`is equal toA. `e^(x)`B. `-(e^(x))/((1+e^(x))^(3))`C. `-(e^(x))/((1+e^(x))^(2))`D. `(1)/((1+e^(x))^(2))`

Answer» Correct Answer - B
Discussion
286.

If`sqrt(1-x^2)+sqrt(1-y^2)=a(x-y),"p r o v et h a t"(dy)/(dx)sqrt((1-y^2)/(1-x^2))`

Answer» Given: `sqrt(1-x^(2))+sqrt(1-y^(2))=a(x-y)." …(i)"`
Putting `x=sin theta and y = sin phi`, it becomes
`cos theta + cos phi =a(sin theta - sin phi)`
`rArr(cos theta+cos phi)/(sin theta-sin phi)=a`
`rArr(2cos((theta+phi)/(2))cos((theta-phi)/(2)))/(2cos((theta+phi)/(2))sin((theta-phi)/(2)))=a`
`rArr cot((theta-phi)/(2))=a rArr theta-phi = 2 cot^(-1)a`
`rArr sin^(-1)x sin^(-1)y=2 cot^(-1)a." ...(ii)"`
On differentiating both sides of (ii) w.r.t. x, we get
`(1)/(sqrt(1-x^(2)))-(1)/(sqrt(1-y^(2))).(dy)/(dx)=0`
Hence, `(dy)/(dx)=sqrt((1-y^(2))/(1-x^(2))).`
Discussion
287.

If `y=(2x+3)^((3x-5))`, find `(dy)/(dx)`.

Answer» Given: `y=(2x+3)^((3x-5))." …(i)"`
Taking logarithm on both sides of (i), we get
`log y = (3x-5)log(2x+3)." …(ii)"`
On differentiating both sides of (ii) w.r.t. x, we get
`(1)/(y).(dy)/(dx)=(3x-5).(d)/(dx){log(2x+3)}+log(2x+3).(d)/(dx)(3x-5)`
`=(3x-5).(1)/((2x+3)).2+log(2x+3).3`
`rArr(dy)/(dx)=y.{((6x-10))/((2x+3))+3log(2x+3)}`
`rArr(dy)/(dx)=(2x+3)^((3x-5)).{((6x-10))/((2x+3))+3log(2x+3)}.`
Discussion
288.

If `y=sqrt((1-cos2x)/(1+cos2x),)x in (0,pi/2)uu(pi/2,pi),`then find `(dy)/(dx)dot`

Answer» We have
`y=sqrt((1-cso2x)/(1+cos2x)),=sqrt((2sin^(2)x)/(2cos^(2)x))=sqrt(tan^(2)x)`
`=|tanx|," where "x in(0,(pi)/(2))cup((pi)/(2),pi)`
`={{:(tan x, x in(0,(pi)/(2)),),(-tan x, x in((pi)/(2),pi),):}`
`therefore" "(dy)/(dx)={{:(sec^(2)x",", x in(0,(pi)/(2)),),(-sec^(2) x",", x in((pi)/(2),pi),):}`
Discussion
289.

The value of `(d)/(dx)10^(mx)` is -A. `10^(mx)log_(e)10`B. `10^(mx)*m log_(e)10`C. `m*10^(mx)`D. none of these

Answer» Correct Answer - B
Discussion
290.

If `y sqrt(1-x^2)+x sqrt(1-y^2)=1`. Prove that `dy/dx=-sqrt((1-y^2)/(1-x^2))`

Answer» We have `xsqrt(1-y^(2))+ysqrt(1-x^(2))=1." …(i)"`
Putting `x = sin theta and y = cos phi` in (i), we get
`sin theta cos phi+cos theta sin phi=1`
`rArr sin(theta+phi)=1`
`rArr (theta+phi)=sin^(-1)(1)`
`rArr sin^(-1)x+sin^(-1)y=(pi)/(2)." ...(ii)"`
On differentiating both sides of (ii) w.r.t. x, we get
`(1)/(sqrt(1-x^(2)))+(1)/(sqrt(1-y^(2))).(dy)/(dx)=0.`
`therefore (dy)/(dx)=-sqrt((1-y^(2))/(1-x^(2))).`
Discussion
291.

`" If "y=sqrt((1-x)/(1+x))," then "(1-x^(2))(dy)/(dx)` is equal toA. `y^(2)`B. `1//y`C. `-y`D. `-y//x`

Answer» Correct Answer - -y
`"We have "y=sqrt((1-x)/(1+x)`
Differentiating w.r.t. x, we get
`(dy)/(dx)=(1)/(2)((1-x)/(1+x))^(1//2-1)(d)/(dx)((1-x)/(1+x))`
`=(1)/(2)sqrt((1+x)/(1-x))cdot((1+x)(-1)-(1-x)(1))/((1+x)^(2))`
`=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))(1-x)^(2)`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1-x)/(1+x))`
`"or "(1-x^(2))(dy)/(dx)=-y`
`"or "(1-x^(2))(dy)/(dx)+y=0`
Discussion
292.

Differentiate log sin x w.r.t. `sqrt(x).`

Answer» `"Let "u=tan^(-1)((sqrt(1+x^(2))-1)/(x))and v= tan^(-1) x.`
`(du)/(dx)=cos x and (dv)/(dx)=-(sin x)/(2sqrt(cos x))`
`"or "(du)/(dv)=(du//dv)/(dv//dx)=(cot x)/(-(sin x)/(2sqrt(cos x)))=-2sqrt(cos x cot x "cosec" x )`
Discussion
293.

If `y^x+x^y+x^x=a^b`, find `(dy)/(dx)`.

Answer» Let `u=x^(x),v=x^(y) and w=y^(x)`. Then,
`y+v+w=a^(b)`
`rArr(du)/(dx)+(dv)/(dx)+(dw)/(dx)=0." …(i) "[because a^(b)=" constant"]`
Now, `u=x^(x)`
`rArr logu=xlog x`
`rArr(1)/(u).(du)/(dx)=x.(d)/(dx)(logx)+logx.(d)/(dx)(x)`
`" [on differentiating both sides w.r.t. x]"`
`rArr(du)/(dx)=u.{x.(1)/(x)+(logx).1}`
`rArr(du)/(dx)=x^(x)(1+logx)." ...(ii)"`
And, `v=x^(y)`
`rArr log v=ylogx`
`rArr(1)/(v).(dv)/(dx)=y.(d)/(dx)(logx)+(logx).(d)/(dx)(y)`
`" [on differentiating both sides w.r.t. x]"`
`rArr(dv)/(dx)=v.[y.(1)/(x)+(logx).(dy)/(dx)]`
`rArr(dv)/(dx)=x^(y){(y)/(x)+(logx)(dy)/(dx)}." ...(iii)"`
And, `w=y^(x)`
`rArr logw=xlogy`
`rArr(1)/(w).(dw)/(dx)=x.(d)/(dx)(logy)+(logy).(d)/(dx)(x)`
`" [on differentiating both sides w.r.t. x]"`
`rArr(dw)/(dx)=w.{x.(1)/(y).(dy)/(dx)+(logy).1}.`
`rArr(dw)/(dx)=y^(x){(x)/(y).(dy)/(dx)+(logy)}." ...(iv)"`
Using (ii), (iii) and (iv) in (i), we get
`x^(x)(1+logx)+x^(y){(y)/(x)+(logx)(dy)/(dx)}+y^(x).{(x)/(y).(dy)/(dx)+(logy)}=0`
`rArr{x^(x)(1+logx)+y.x^((y-1))+y^(x)(logy)}+{x^(y)(logx)+xy^((x-1))}(dy)/(dx)=0`
`rArr(dy)/(dx)=(-{x^(x)(1+logx)+y.x^((y-1))+y^(x)(logy)})/({x^(y)(logx)+xy^((x-1))}).`
Discussion
294.

`" If "y=sqrt((1-x)/(1+x))," then "(1=x^(2))(dy)/(dx)` is equal toA. `y^(2)`B. `1//y`C. `-y`D. `-y//x`

Answer» `"We have "y=sqrt((1-x)/(1+x)`
Differentiating w.r.t. x, we get
`(dy)/(dx)=(1)/(2)((1-x)/(1+x))^(1//2-1)(d)/(dx)((1-x)/(1+x))`
`=(1)/(2)sqrt((1+x)/(1-x))cdot((1+x)(-1)-(1-x)(1))/((1+x)^(2))`
`=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))(1-x)^(2)`
`"or "(1-x)^(2)(dy)/(dx)=-sqrt((1-x)/(1+x))`
`"or "(1-x^(2))(dy)/(dx)=-y`
`"or "(1-x^(2))(dy)/(dx)+y=0`
Discussion
295.

If `sin^(-1)((x^2-y^2)/(x^2+y^2))=loga ,t h e n(dy)/(dx)`is equal to`x/y`(b) `y/(x^2)``(x^2-y^2)/(x^2+y^2)`(d) `y/x`A. `(x)/(y)`B. `(y)/(x^(2))`C. `(x^(2)-y^(2))/(x^(2)+y^(2))`D. `(y)/(x)`

Answer» `"We have "sin^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=log a`
`"or "(x^(2)-y^(2))/(x^(2)+y^(2))=sin (log a)`
`"or "(1-tan^(2)theta)/(1+tan^(2)theta)=sin(log a)" "("on putting "y= x tan theta)`
`"or "cos 2theta= sin (log a)`
`"or "2theta=cos^(-1)(sin (log a))`
`"or "theta=(1)/(2)cos^(-1)(sin (log a))`
`"or "tan^(-1)((y)/(x))=(1)/(2)cos^(-1)(sin (log a))`
`"or "(y)/(x)=tan ((1)/(2)cos^(-1)(sin (loga )))`
Differentiating w.r.t. x, we get
`(x(dy)/(dx)-y)/(x^(2))=0`
`"or "x(dy)/(dx)-y=0`
`"or "(dy)/(dx)=(y)/(x)`
Discussion
296.

If `y=x^x^x^^((((oo))))`, find `(dy)/(dx)dot`

Answer» `y=x^(y)`
`"or "log y = y log x`
`"or "(1)/(y)(dy)/(dx)=(dy)/(dx)xxlog x+y(1)/(x)" "[Diff. both sides w.r.t.x]`
`"or "(dy)/(dx)({1-y log x })/(y)=(y)/(x)`
`"or "(dy)/(dx)=(y^(2))/(x(1-y log x))`
Discussion
297.

If `x^y+y^x=a^b` , then find `dy/dx`.

Answer» Let `u=x^(y) and v=y^(x)`.
Then, `u+v=a^(b)rArr(du)/(dx)+(dv)/(dx)=0." ...(i) "[becausea^(b)=" constant"]`
Now, `u=x^(y)rArrlogu=ylogx" [taking log on both sides]"`
`rArr(1)/(u).(du)/(dx)=y.(1)/(x)+logx.(dy)/(dx)" [on differentiaton]"`
`rArr(du)/(dx)=u[(y)/(x)+logx.(dy)/(dx)]`
`rArr(du)/(dx)=x^(y)[(y+xlogx.(dy)/(dx))/(x)]`
`rArr(du)/(dx)=x^(y-1)[y+xlogx.(dy)/(dx)].`
And, `v=y^(x)rArr log v=x logy" [taking log on both sides]"`
`rArr(1)/(v).(dv)/(dx)=x.(1)/(y).(dy)/(dx)+logy" [on differentiation]"`
`rArr(dv)/(dx)=v.[(x)/(y).(dy)/(dx)+logy]rArr(dv)/(dx)=y^(x){(x.(dy)/(dx)+ylogy)/(y)}`
`rArr(dv)/(dx)=y^((x-1)).{x.(dy)/(dx)+ylogy}.`
Using (i), we get `(du)/(dx)+(dv)/(dx)=0`
`rArrx^((y-1)){y+xlogx.(dy)/(dx)}+y^((x-1)).{x(dy)/(dx)+ylogy}=0`
`rArr{x^(y)(logx)+x.y^((x-1))}.(dy)/(dx)=-{y.x^((y-1))+y^(x)(logy)}.`
`therefore(dy)/(dx)=(-{y.x^((y-1))+y^(x)(logy)})/({x^(y)(logx)+xy^((x-1))}).`
Discussion
298.

If`y=(sqrt(a+x)-sqrt(a-x))/(sqrt(a+x)+sqrt(a-x)),t h e n(dy)/(dx)i se q u a lto``(a y)/(xsqrt(a^2-x^2))`(b) `(a y)/(sqrt(a^2-x^2))``(a y)/(xsqrt(a^2-x^2))`(d) none of theseA. `(ay)/(xsqrt(a^(2)-x^(2)))`B. `(ay)/(sqrt(a^(2)-x^(2)))`C. `(ay)/(xsqrt(x^(2)-a^(2)))`D. none of these

Answer» `y=(sqrt(a+x)-sqrt(a-x))/(sqrt(a+x)+sqrt(a-x))`
`=((sqrt(a+x)-sqrt(a-x))^(2))/((a+x)-(a-x))`
`=((a+x)+(a-x)-2(sqrt(a^(2)-x^(2))))/(2x)`
`=(2a-2sqrt(a^(2)-x^(2)))/(2x)=(a-sqrt(a^(2)-x^(2)))/(x)" (1)"`
Differentiating w.r.t. x, we get
`(dy)/(dx)=(x[-(1)/(2sqrt(a^(2)-x^(2)))(-2x)]-(a-sqrt(a^(2)-x^(2))))/(x^(2))`
`=(x^(2)-asqrt(a^(2)-x^(2)+a^(2)-x^(2)))/(x^(2)sqrt(a^(2)-x^(2)))=(a(a-sqrt(a^(2)-x^(2))))/(x^(2)sqrt(a^(2)-x^(2)))`
`(a)/(xsqrt(a^(2)-x^(2)))[(a-sqrt(a^(2)-x^(2)))/(x)]=(ay)/(xsqrt(a^(2)-x^(2)))" [by (1)]"`
Discussion
299.

`xsqrt(1+y)+ysqrt(1+x)=0` then `(dy)/(dx)=`A. `(1)/((1+x)^(2))`B. `-(1)/((1+x)^(2))`C. `(1)/(1+x^(2))`D. `(1)/(1-x^(2))`

Answer» Correct Answer - B
Discussion
300.

If `x^(y)=e^(x-y)`, prove that `(dy)/(dx)=(logx)/((1+logx)^(2)).`

Answer» We have
`x^(y)=e^(x-y)rArr ylog x=(x-y)`
`rArr (1+logx)y=x`
`rArry=(x)/((1+logx))." …(i)"`
On differentiating both sides of (i) w.r.t. x, we get
`(dy)/(dx)=((1+logx).(d)/(dx)(x)-x.(d)/(dx)(1+logx))/((1+logx)^(2))`
`=((1+logx).1-x.(1)/(x))/((1+logx)^(2))=((1+logx-1))/((1+logx)^(2))=(logx)/((1+logx)^(2)).`
Discussion