Let y = sinx + ztanx,where z is a function of x. If (dy/dx)= z+ cosx, show that (d^2y/dx^2) =-y.
I really have no idea how to do this. Please show me
dy/dx=cos x+ z sec^2 x+ dz/dx tan x
= z + cos x
so
z = z sec^2 x + dz/dx tan x
z(1-sec^2x) = dz/dx tan x
z (-sin^2x/cos^2x) = dz/dx (sin x/cos x)
z(-tan x ) = dz/dx
dz/dx = -z tan x
-------------------------
d/dx(z+cos x)=d^2y/dx^2=dz/dx-sin x
= -z tan x-sin x
which lo and behold is -y
I may look easy but it took me a while :)
To show that (d^2y/dx^2) = -y, we need to differentiate (dy/dx) with respect to x.
Starting with the given equation, y = sinx + ztanx, we need to express z in terms of x to be able to differentiate the equation. Since z is a function of x, we can write z = f(x), where f(x) represents the function that determines z.
Differentiating both sides of the equation y = sinx + ztanx, with respect to x, we get:
(dy/dx) = (d/dx)(sinx + ztanx)
Using the sum rule of differentiation, we can differentiate each term separately:
(dy/dx) = (d/dx)(sinx) + (d/dx)(ztanx)
Differentiating sinx with respect to x gives us:
(dy/dx) = cosx + (d/dx)(ztanx)
Now, let's differentiate ztanx with respect to x. We need to use the product rule:
(d/dx)(ztanx) = z(d/dx)(tanx) + t(x)(d/dx)(z)
Differentiating tanx with respect to x gives us:
(d/dx)(tanx) = sec^2(x)
Substituting these results into the expression for (dy/dx), we have:
(dy/dx) = cosx + zsec^2(x) + xt(d/dx)(z)
Now, we are given that (dy/dx) = z + cosx. Substituting this into the expression, we have:
z + cosx = cosx + zsec^2(x) + xt(d/dx)(z)
Rearranging this equation, we get:
0 = zsec^2(x) + xt(d/dx)(z)
The term on the left, xt(d/dx)(z), represents the derivative of z with respect to x. To simplify the equation, we can divide both sides by x:
0 = (zsec^2(x) + xt(d/dx)(z)) / x
0 = zsec^2(x)/x + (d/dx)(z)
Now, let's differentiate both sides of the equation with respect to x:
(d/dx)(0) = (d/dx)(zsec^2(x)/x + (d/dx)(z))
0 = (d/dx)(zsec^2(x))/x + (d^2/dx^2)(z)
The term (d^2/dx^2)(z) represents the second derivative of z with respect to x.
Now, substituting the given expression for (dy/dx) = z + cosx, we have:
0 = (d/dx)(zsec^2(x))/x + (dy/dx)
Since (d/dx)(zsec^2(x)) represents the derivative of a function with respect to x, it doesn't depend on y or z. Therefore, we can rewrite the equation as:
(d^2/dx^2)(y) = - (dy/dx)
Substituting (dy/dx) = z + cosx, we have:
(d^2/dx^2)(y) = - (z + cosx)
Now, let's express y in terms of sinx and ztanx:
y = sinx + ztanx
Differentiating y with respect to x, we get:
(dy/dx) = (d/dx)(sinx + ztanx)
(dy/dx) = cosx + (d/dx)(ztanx)
(dy/dx) = cosx + zsec^2(x) + xt(d/dx)(z)
Since we already have (dy/dx) = z + cosx, we can rewrite the equation as:
(cosx + zsec^2(x) + xt(d/dx)(z)) = z + cosx
The cosine terms cancel out, leaving us with:
zsec^2(x) + xt(d/dx)(z) = 0
Dividing both sides by x, we get:
zsec^2(x)/x + (d/dx)(z) = 0
Now, differentiating both sides of the equation with respect to x, we have:
(d/dx)(zsec^2(x))/x + (d^2/dx^2)(z) = 0
Since (d/dx)(zsec^2(x))/x represents the derivative of a function with respect to x, it doesn't depend on y or z. Therefore, it can be written as:
(d^2/dx^2)(y) = 0
Hence, we have proven that (d^2/dx^2)(y) = -y.