find equations for the tangent and the normal at P (1n2, 2k) on the curve with equation y=ke^x, where k is a constant.

(b) given that y=sin^-1x, show that (1-x^2)d^2y/dx^2-xdy/dx=0

To find the equations for the tangent and the normal at point P (1n2, 2k) on the curve with equation y = ke^x, we need to find the first derivative of the curve, evaluate it at point P, and then find the equation of the tangent and normal lines using the derivative.

(a) Finding the derivative of y = ke^x:
To find the first derivative, we differentiate y = ke^x with respect to x using the chain rule.
dy/dx = d/dx (ke^x)

The derivative of ke^x with respect to x is ke^x, since e^x is the derivative of x with respect to x.
So, dy/dx = ke^x.

Now, we can evaluate this derivative at point P (1n2, 2k):
dy/dx = k*e^(1n2)
dy/dx = k*e^(ln(2))
dy/dx = k*2
dy/dx = 2k

So, the slope of the tangent at point P is 2k.

The equation for a straight line (tangent) is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.

Using point P (1n2, 2k), the equation for the tangent is:
y - 2k = 2k(x - 1n2)
y - 2k = 2k(x - 1)
y = 2kx - 2k + 2k = 2kx

So, the equation for the tangent at point P is y = 2kx.

To find the equation for the normal, the slope of the normal line is the negative reciprocal of the slope of the tangent. The negative reciprocal of 2k is -1/(2k).

Therefore, the equation for the normal at point P is:
y - 2k = -1/(2k)(x - 1n2)
y - 2k = -1/(2k)x + 1/2
y = -1/(2k)x + 1/2 + 2k = -1/(2k)x + (4k + 1)/2

So, the equation for the normal at point P is y = -1/(2k)x + (4k + 1)/2.

(b) Given y = sin^(-1)(x), we need to find the second derivative (d²y/dx²) and first derivative (dy/dx) and then show that (1 - x²)d²y/dx² - x(dy/dx) = 0.

First, let's find the first derivative dy/dx and the second derivative d²y/dx² of y = sin^(-1)(x).

Using the chain rule, we differentiate y = sin^(-1)(x) as follows:
dy/dx = d/dx[sin^(-1)(x)]

To apply the chain rule, we let u = sin^(-1)(x), then differentiate u with respect to x:
du/dx = 1/sqrt(1 - x²)

Now, we differentiate y = sin^(-1)(x) with respect to x using the chain rule:
dy/dx = du/dx * d/dx(sinx)
dy/dx = (1/sqrt(1 - x²)) * cos(x)

Next, let's find the second derivative d²y/dx²:
d²y/dx² = d/dx(dy/dx)
d²y/dx² = d/dx[(1/sqrt(1 - x²)) * cos(x)]

To differentiate the product of two functions, we use the product rule:
d²y/dx² = [(1/sqrt(1 - x²)) * d(cos(x))/dx] + [cos(x) * d/dx(1/sqrt(1 - x²))]

The derivative of cos(x) with respect to x is -sin(x), and the derivative of (1/sqrt(1 - x²)) with respect to x can be found using the quotient rule:
d/dx(1/sqrt(1 - x²)) = [(0 * sqrt(1 - x²)) - (1 * d(sqrt(1 - x²))/dx)] / (sqrt(1 - x²))^2
d/dx(1/sqrt(1 - x²)) = -d(sqrt(1 - x²))/dx / (1 - x²)

The derivative of sqrt(1 - x²) with respect to x can be found using the chain rule:
d(sqrt(1 - x²))/dx = (1/2) * (1 - x²)^(-1/2) * (-2x)
d(sqrt(1 - x²))/dx = -x / sqrt(1 - x²)

Now, substituting the derivatives into d²y/dx²:
d²y/dx² = [(1/sqrt(1 - x²)) * (-sin(x))] + [cos(x) * (-x/sqrt(1 - x²))]

Simplifying further:
d²y/dx² = -sin(x)/sqrt(1 - x²) - xcos(x)/sqrt(1 - x²)
d²y/dx² = -(sin(x) + xcos(x))/sqrt(1 - x²)

Now, let's substitute the first derivative dy/dx and the second derivative d²y/dx² into the given expression (1 - x²)d²y/dx² - x(dy/dx):
(1 - x²)d²y/dx² - x(dy/dx) = (1 - x²)(-(sin(x) + xcos(x))/sqrt(1 - x²)) - x((1/sqrt(1 - x²)) * cos(x))

Simplifying further:
(1 - x²)d²y/dx² - x(dy/dx) = -(sin(x) + xcos(x)) - xcos(x)

Finally, simplifying the expression:
(1 - x²)d²y/dx² - x(dy/dx) = -sin(x) - xcos(x) - xcos(x)
(1 - x²)d²y/dx² - x(dy/dx) = -sin(x) - 2xcos(x)

We can see that the expression (1 - x²)d²y/dx² - x(dy/dx) is equal to -sin(x) - 2xcos(x), not zero.

Hence, we have shown that (1 - x²)d²y/dx² - x(dy/dx) does not equal zero for y = sin^(-1)(x).