NEET Physics Class 11 Chapter 10 Mathematical Tools – Rules For Differentiation
Derivative Of A Constant:
The first rule of differentiation is that the derivative of every constant function is zero. If c is constant, then \(\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{c}\) = 0
⇒ \(\frac{d}{d x}(8)=0\)
⇒ \(\frac{d}{d x}\left(-\frac{1}{2}\right)=0\)
Power Rule:
If n is a real number, then \(\frac{d}{d x} x^n=n x^{n-1}\)
To apply the power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.
⇒ \(\begin{array}{l|l|l|l|l|l}\mathrm{f} & \mathrm{x} & \mathrm{x}^2 & x^3 & x^4 & \ldots \\\hline f^{\prime} & 1 & 2 x & 3 x^2 & 4 x^3 & \ldots\end{array}\)
Question 1.
1. \(\frac{d}{d x}\left(\frac{1}{x}\right)\)
⇒ \(\frac{d}{d x}\left(x^{-1}\right)\)
⇒ \((-1) x^{-2}\)
⇒ \(-\frac{1}{x^2}\)
2. \(\frac{d}{d x}\left(\frac{4}{x^3}\right)\)
⇒ \(4 \frac{d}{d x} \quad\left(x^{-3}\right)\)
⇒ \(4(-3) x^{-4}\)
⇒ \(-\frac{12}{x^4}\)
The Constant Multiple Rule:
If u is a differentiable function of x, and c is a constant, then \(\frac{d}{d x}(c u)=c \frac{d u}{d x}\)
In particular, if n is a positive integer, then \(\frac{d}{d x}\left(c x^n\right)=c n x^{n-1}\)
Question 2. The derivative formula
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^2\right)\)
= 3(2x)
= 6x
says that if we rescale the graph of y = x2 by multiplying each y-coordinate by 3, then we multiply the slope at each point by 3.
Question 3. A useful special case
The derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c = –1 gives.
⇒ \(\frac{d}{d x}(-u)=\frac{d}{d x}(-1 \cdot u)=-1 \quad \frac{d}{d x}(u)=-\frac{d}{d x}\)
The Sum Rule
The derivative of the sum of two differentiable functions is the sum of their derivatives. If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v are both differentiable functions in their derivatives.
⇒ \(\frac{d}{d x}(u-v)=\frac{d}{d x}[u+(-1) v]\)
⇒ \(\frac{d u}{d x}+(-1) \frac{d v}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\)
The Sum Rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u1, u2,………un are differentiable at x, then so is u1+ u2+ ……..+ un, and
⇒ \(\frac{d}{d x}\left(u_1+u_2+\ldots . .+u_n\right)\)
⇒ \(\frac{d u_1}{d x}+\frac{d u_2}{d x}+\ldots \ldots .+\frac{d u_n}{d x}\)
Question 4.
1. \(y= x^4+12 x \\\)
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(x^4\right)+\frac{d}{d x}(12 x)\)
⇒ \(4 x^3+12\)
2. \(y =x^3+\frac{4}{3} x^2-5 x+1\)
⇒ \(\frac{d y}{d x} =\frac{d}{d x}\left(x^3\right)+\frac{d}{d x}\left(\frac{4}{3} x^2\right)-\frac{d}{d x}(5 x)+\frac{d}{d x}(1)\)
⇒ \(3 x^2+\frac{4}{3} \cdot 2 x-5+0\)
⇒ \(3 x^2+\frac{8}{3} x-5\)
Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in the above example.
The Product Rule
If u and v are differentiable at x, then so is their product uv, and \(\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\)
The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation (uv)’ = uv’ + vu’.
While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance,
⇒ \(\frac{d}{d x}(x . x)=\frac{d}{d x}\left(x^2\right)=2 x,\),
while \(\frac{\mathrm{d}}{\mathrm{dx}} \text { (x) } \frac{\mathrm{d}}{\mathrm{dx}} \text {.(x) }\)= 1.1 = 1.
Question 5. Find the derivatives of y = (x2 + 1) (x3 + 3).
Answer:
From the product Rule with u = x2 + 1 and v = x3 + 3, we find
⇒ \(\frac{d}{d x}\left[\left(x^2+1\right)\left(x^3+3\right)\right]\)
= (x2 + 1) (3x2) + (x3 + 3) (2x)
= 3x4 + 3x2 + 2x4 + 6x
= 5x4 + 3x2 + 6x.
For example can be done as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial. We now check :
y=(x2 + 1)(x3 + 3)=x5 +x3+3x2 + 3
⇒ \(\frac{d y}{d x}=5 x^4+3 x^2+6 x\)
This is in agreement with our first calculation.
There are times, however, when the product Rule must be used. In the following examples. We have only numerical values to work with.
Question 6. Let y = uv be the product of the functions u and v. Find y’ (2)if u’(2)= 3, u’(2)= –4, v(2)= 1, and v’(2)= 2.
Answer:
From the Product Rule, in the form
y’ = (uv)’ = uv’ + vu’
we have y’(2) = u(2) vs(2) + v(2) up(2)
= (3)(2)+(1) (–4)
= 6 – 4 = 2.
The Quotient Rule
If u and v are differentiable at x, and v(x) ≠ 0, then the quotient u/v is differentiable at x, and
⇒ \(\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2}\)
Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives.
Question 7. Find the derivative of y = \(\frac{t^2-1}{t^2+1}\)
Answer: We apply the Quotient Rule with u = t2 – 1 and v = t2 + 1:
⇒ \(\frac{d y}{d t}=\frac{\left(t^2+1\right) \cdot 2 t-\left(t^2-1\right) \cdot 2 t}{\left(t^2+1\right)^2}\)
⇒ \(\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v}(\mathrm{du} / \mathrm{dt})-\mathrm{u}(\mathrm{dv} / \mathrm{dt})}{\mathrm{v}^2}\)
⇒ \(\frac{2 t^3+2 t-2 t^3+2 t}{\left(t^2+1\right)^2}=\frac{4 t}{\left(t^2+1\right)^2}\)
Derivative Of Sine Function
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sinx) cos x
Question 8.
1. y = x2 – sin x :
⇒ \(\frac{d y}{d x}=2 x-\frac{d}{d x}(\sin x)\) Difference Rule
= 2x -cos x
2. y = x2 sin x :
⇒ \(\frac{d y}{d x}=x^2 \frac{d}{d x}(\sin x)+2 x \sin x\)Product Rule
⇒ \(x^2 \cos x+2 x \sin x\)
3. y = \(\frac{\sin x}{x}\)
⇒ \(\frac{d y}{d x}=\frac{x \cdot \frac{d}{d x}(\sin x)-\sin x .1}{x^2}\) Quotient rule
⇒ \(\frac{x \cos x-\sin x}{x^2}\)
Derivative Of Cosine Function
⇒ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(cosx)= – sinx
Question 9.
1. y = 5x + cos x
⇒ \(\frac{d y}{d x}=\frac{d}{d x}(5 x)+\frac{d}{d x}(\cos x)\) Sum Rule
= 5 – sin x
2. y = sinx cosx
⇒ \(\frac{d y}{d x}=\sin x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(\sin x)\) Product Rule
= sinx (– sinx) + cosx (cosx)
= cos2 x – sin2 x
Derivatives Of Other Trigonometric Functions
Because sin x and cos x are differentiable functions of x, the related functions
tan x = \(\frac{\sin x}{\cos x}\)
cot x = \(\frac{\cos x}{\sin x}\)
sec x = \(\frac{1}{\cos x}\)
cosec x = \(\frac{1}{\sin x}\)
are differentiable at every value of x at which they are defined. There derivatives. Calculated from the Quotient Rule, are given by the following formulas.
⇒ \(\frac{d}{d x}\)(tan x) = sec2 x ;
⇒ \(\frac{d}{d x}\)(sec x) = sec x tan
⇒ \(\frac{d}{d x}\)(cot x) = – cosec2 x ;
⇒ \(\frac{d}{d x}\)(cosec x) = – cosec x cot x
Question 10. Find dy / dx if y = tan x.
Answer:
⇒ \(\frac{d}{d x}(\tan x)\)
⇒ \(\frac{d}{d x}\left(\frac{\sin x}{\cos x}\right)\)
⇒ \(\frac{\cos x \frac{d}{d x}(\sin x)-\sin x \frac{d}{d x}(\cos x)}{\cos ^2 x}\)
⇒ \(\frac{\cos x \cos x-\sin x(-\sin x)}{\cos ^2 x}\)
⇒ \(\frac{\cos ^2 x+\sin ^2 x}{\cos ^2 x}\)
⇒ \(\frac{1}{\cos ^2 x}=\sec ^2 x\)
Question 11.
1. \(\frac{d}{d x}(3 x+\cot x)\)
⇒ \(3+\frac{d}{d x}(\cot x)\)
⇒ \(3-{cosec}^2 x\)
2. \(\frac{d}{d x}\left(\frac{2}{\sin x}\right)\)
⇒ \(\frac{d}{d x}(2{cosec} x)\)
⇒ \(2 \frac{d}{d x}({cosec} x)\)
= 2 (–cosec x cot x)
= – 2 cosec x cot x
Derivative Of Logarithm And Exponential Functions
⇒ \(\frac{d}{d x}\left(\log _e x\right)=\frac{1}{x}\)
⇒ \(\frac{d}{d x}\left(e^x\right)=e^x\)
Question 12. y=\(e^x \cdot \log _e(x)\)
Answer:
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(e^x\right) \cdot \log (x)+\frac{d}{d x}\left[\log _e(x)\right] e^x\)
⇒ \(\frac{d y}{d x}=e^x \cdot \log _e(x)+\frac{e^x}{x}\)
Question 13. \(\frac{d}{d t}\)
Answer: ω cos ωt
Question 14. \(\frac{d}{d t}\)(cos t)
Answer: −ω sin ωt
Question 15.
1. \(\frac{d}{d x} \cos 3 x \quad\)
⇒ \(-\sin 3 x \frac{d}{d x} 3 x\)
= – sin 3x
2. \(\frac{d}{d x} \sin 2 x \quad\)
⇒ \(\cos 2 x \frac{d}{d x}(2 x)\)= cos 2x.2
= 2 cos 2x
3. \(\frac{d}{d t}\)(A sin (ωt + φ)
= A cos (ωt + φ)\(\frac{d}{d t}\)(ωt + φ)
= A cos (ωt + φ). ω.
= A ω cos (ωt + φ)
Question 16. \(\frac{d}{d x}\left(\frac{1}{3 x-2}\right)\)
Answer:
⇒ \(\frac{d}{d x}(3 x-2)^{-1} \quad\)
⇒ \(-1(3 x-2)^{-2} \frac{d}{d x}(3 x-2)\)
⇒ \(-1(3 x-2)^{-2}(3)\)
⇒ \(-\frac{3}{(3 x-2)^2}\)
Question 17. \(\frac{d}{d t}[A \cos (\omega t+\varphi)]\)
Answer:
= –Aω sin (ωt + φ)
Chain Rule
If f (x) is given as function of g(x) i.e., y = f(g(x)) and we are required to find \(\frac{d y}{d x}\) assume dx g(x)= u
⇒ y = f
⇒ \(\frac{d y}{d u}=f^{\prime}(4)\)
⇒ \(\frac{d u}{d x}=g^{\prime}(x)\)
Example: y = sin (x2)
y = log(x2 + 5x)
y = sin (cos x)
y = A sin (ωt + φ), A,ω,φ, are constant
Radian Vs. Degrees
⇒ \(\frac{d}{d x} \sin \left(x^{\circ}\right)=\frac{d}{d x} \sin \left(\frac{\pi x}{180}\right)\)
⇒ \(\frac{\pi}{180} \cos \left(\frac{\pi x}{180}\right)=\frac{\pi}{180} \cos \left(x^{\circ}\right)\)