Theorems and Laws [1]
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
Given, y = 5 cos x – 3 sin x
Differentiating both sides with respect to x,
`dy/dx = 5 d/dx cos x - 3 d/dx sin x`
= 5 (−sin x) − 3 cos x
= −5 sin x − 3 cos x
Differentiating both sides again with respect to x,
`(d^2 y)/dx = - 5 d/dx sin x - 3 d/dx cos x`
= −5 cos x − 3 (−sin x)
= 3 sin x − 5 cos x
Hence, `(d^2 y)/dx^2 + y` = 0
(3 sin x − 5 cos x) + (5 cos x − 3 sin x) = 0 ...(On substituting the value of y)
Key Points
If x = f(t) and y = g(t) are differentiable functions of parameter t, then
\[\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)},\frac{\mathrm{d}x}{\mathrm{d}t}\neq0\]
- If an equation contains both x and y and cannot be solved directly for y, it is called an implicit function.
- Implicit functions are generally written in the form:
f(x, y) = 0 - To differentiate an implicit function, differentiate both sides with respect to x, treating y as a function of x.
First Derivative Test:
Maxima at x = c:
- f′(c) = 0
- f′(c − h) > 0 and f′(c + h) < 0
Minima at x = c:
- f′(c) = 0
- f′(c − h) < 0 and f′(c + h) > 0
Second Derivative Test:
- If f′(a) = 0 and f″(a) < 0 → Maximum
- If f′(a) = 0 and f″(a) > 0 → Minimum
- If f″(a) = 0 → Test fails (use first derivative)
