Solution:Â
Firstly taking sin on both sides, hence we get x = siny this equation is nothing but a function of y. Instead of finding dy/dx we will find dx/dy, so by definition of derivative we can write ((f(y + h) - f(y))/h), where h -> 0 under the limiting condition (see fourth line). Now replace the function with ((sin(y + h) - siny)/h) where h -> 0 under the limiting condition. Using the identity we can solve further. As we see in the last line of the below solution that siny and cosy are not dependent on the limit h -> 0 that's why we had taken them out.
sin(sin x) = sin y
x = sin y
By definition of derivative,
dx/dy = limh->0 {f(y + h) - f(y)} / h
= limh->0 { sin(y + h) - siny } / h
Using identity: sin(A + B) = sinA.cosB + cosA.sinB, we can write,
= limh->0 (sin y . cos h + cos y . sin h - sin y) / h
= limh->0 (sin y . cos h - sin y + cos y . sin h) / h
= limh->0 {sin y(cos h - 1) / h} + {cos y . sin h) / h}
= sin y. limh->0 {(cos h - 1) / h} + cos y. limh->0 {sin h / h}
Now, we had taken -1 common from the expression (cos h-1) and we get (see in 1st line of below figure). Now using the formula as written in line 2 of the below figure we can write our expression dx/dy = cos y, if we reciprocal this term we get dy/dx = 1/cos y this. We know that sin2 x + cos2 x = 1, by simplifying this formula to get our answer, we simplified it till the 6th line of the below figure.Â
But how had we written the final answer to this problem?
Since -pi/2 ≤ sin-1x ≤ pi/2. Hence -pi/2 ≤ y ≤ pi/2, we had written y in place of sin-1x, look at above figure second line we had written x = siny, if we write this for y we can write this like y = sin-1x this, that's why we had written y in place of sin-1x. This implies 0 ≤ cosy ≤ 1 because y is an angle which lies first and fourth quadrant only, but one thing to note here, since cosy is in the denominator of dy/dx hence it cannot be zero,Â
Now we remove the equality 0 < cos y ≤ 1 by this inequality we can clearly say that cosy is a positive property, hence we can remove -ve sign from the second last line of the below figure. So, this implies dy/dx = 1 over the quantity square root of (1 - x2), which is our required answer.
Note: In the solution after removing square we are getting square-root on another side and with square-root +ve and - ve both signs take place which is denoted by +-squareroot in the solution.
= sin y. limh->0 { (cos h - 1) / h } + cos y. limh->0 { sin h / h }
By using the formula: limh->0 (1 - cos h) / h = 0 and limh->0 sin h / h = 1, we can write,
dx / dy = cos y
dy / dx = 1 / cos y .....(1)
We know that sin2y + cos2y = 1, so cos2y = 1 - sin2y
⇒ cos = +- √(1 - sin2y), taking siny = x
we get, cosy = +-√(1 - x2)
dy / dx = 1 / √(1 - x2)
Solution:Â
For solving and finding the cos-1x ,we have to remember below three listed formulae.
- limh->0 {f(x + h) - f(x)} / h
- cos-1x + sin-1x = pi/2
- cos-1x = pi/2 - sin-1x
Now, let's solve, we have.
f(x) = cos-1x
f(x + h) = cos-1(x + h)
limh->0 {cos-1(x + h ) - cos-1(x)} / h
limh->0 {pi/2 - sin-1(x + h) - (pi/2 - sin-1x) } / h
limh->0 {pi/2 - sin-1(x + h) - pi/2 + sin-1x } / h
Taking - sign common, we get
- limh->0 {sin-1(x + h) - sin-1x} / h
Since we know that limh->0 { sin-1(x + h) - sin-1x } / h = 1 / √(1 - x2)
Putting the value in our solution we get,
- 1 / √(1 - x2)
Solution:Â
For solving and finding tan-1x, we have to remember some formulae, listed below.
- limh->0 {f(x + h) - f(x)} / h
- tan-1(θ/θ) = 1
- tan-1x - tan-1y = tan-1[(x - y) / (1 + xy)]
f(x) = tan-1x
f(x + h) = tan-1(x + h)
Apply 1st formula
limh->0 {tan-1(x + h) - tan-1x } / h
Now Apply 3rd formula
limh->0 tan-1[(x - h - x) / (1 + (x + h)x] / h
limh->0 tan-1[(h  / (1 + x2 + xh ] / h . [(1 + x2 + xh) / (1 + x2 + xh)]
limh->0 tan-1 {h / 1 + x2 + xh} / {h / 1 + x2 + xh} . limh->0 1 / 1 + x2 + xh
Now we made the solution like so that we apply the 2nd formula
= 1 . 1 / (1 + x2 + x . 0)
= 1 / (1 + x2)
Solution:
As we had solved the first problem in the same way we are going to solve this problem too, we have to find out the derivative of the above question, so first, we have to substitute the formulae of tan-1x as we discuss in the above list (line 3). Then apply the chain rule. As we see 1/a is constant, so we take it out and applying the chain rule in tan-1(x/a). Solved it by taking the derivative after applying chain rule.
By using chain rule,
y' = ((1 / a) tan-1(x / a))'
= (1 / a) {1 / (1 + (x / a))} . (x / a)'
= 1 / a . {1 / (1+ (x2 / a2))} . (1 / a)
= 1 / a2 . {a / (a2 + x2)}
= a / a2 + x2
from eq (1), formula of cos(x) = base / hyp , we can find the perpendicular of triangle
perpendicular = √(1 - x2)
formula of cosec(x) = hyp / perpendicular, which is,
= 1 / √(1 - x2)
Putting the value of cosec in eq(2), we get
y' = -1 / √(1 - x2)  Â