Find the derivative of x^2-2 at x=10.
Hint. Apply the power rule to the x^2 term (the constant -2 contributes 0), then substitute x=10.
Using the power rule on x^2 gives 2x, and the constant term -2 differentiates to 0, so d/dx(x^2-2)=2x. Evaluating this derivative function at x=10: 2(10)=20.
✦ Working through each part gives: 20.
Find the derivative of x at x=1.
Hint. The derivative of x with respect to itself is the constant function 1, from the power rule with n=1.
By the power rule, d/dx(x^1)=1.x^0=1, a constant function. This value is the same at every point, including at x=1.
✦ Working through each part gives: 1.
Find the derivative of 99x at x=100.
Hint. Apply the power rule (or the constant-multiple rule) to 99x.
d/dx(99x)=99. This is the same at every point, including x=100.
✦ Working through each part gives: 99.
Find the derivative of the following functions from first principles: (i) x^3-27 (ii) (x-1)(x-2) (iii) 1/x^2 (iv) (x+1)/(x-1).
Hint. Set up f'(x)=lim(h->0)[f(x+h)-f(x)]/h for each function and simplify the difference quotient before taking the limit.
(i) f(x+h)-f(x) = (x+h)^3-x^3 = 3x^2h+3xh^2+h^3. Dividing by h and taking h->0: 3x^2. (ii) f(x)=x^2-3x+2, so f(x+h)-f(x)=2xh+h^2-3h. Dividing by h and taking h->0: 2x-3. (iii) f(x+h)-f(x)=1/(x+h)^2-1/x^2 = [x^2-(x+h)^2]/[x^2(x+h)^2] = [-2xh-h^2]/[x^2(x+h)^2]. Dividing by h and taking h->0: -2x/x^4=-2/x^3. (iv) f(x+h)-f(x) = (x+h+1)/(x+h-1) - (x+1)/(x-1); combining over a common denominator, the numerator simplifies to -2h, so dividing by h and taking h->0 (with denominator (x-1)^2): -2/(x-1)^2.
✦ Working through each part gives: (i) 3x^2 (ii) 2x-3 (iii) -2/x^3 (iv) -2/(x-1)^2.
For the function f(x) = x^100/100 + x^99/99 + ... + x^2/2 + x + 1, prove that f'(1) = 100.f'(0).
Hint. Differentiate termwise using the power rule, then evaluate at x=1 and at x=0 separately.
Differentiating termwise: f'(x) = x^99 + x^98 + ... + x + 1 (100 terms, since d/dx(x^k/k)=x^(k-1) for each k from 1 to 100). At x=1: f'(1) = 1+1+...+1 (100 ones) = 100. At x=0: every term x^(k-1) is 0 except the constant term (k=1, giving x^0=1), so f'(0)=1.
✦ Working through each part gives: f'(1)=100 and f'(0)=1, so f'(1) = 100.f'(0), as required.
Find the derivative of x^n + ax^(n-1) + a^2x^(n-2) + ... + a^(n-1)x + a^n for some fixed real number a.
Hint. Differentiate termwise using the power rule; the last term a^n is a constant.
This is a sum of terms a^k.x^(n-k) for k=0 to n. Differentiating each (for k<n): d/dx[a^k.x^(n-k)] = a^k(n-k)x^(n-k-1). The final term a^n (k=n) is constant and differentiates to 0.
✦ Working through each part gives: n.x^(n-1) + (n-1)a.x^(n-2) + (n-2)a^2.x^(n-3) + ... + 2a^(n-2)x + a^(n-1).
For some constants a and b, find the derivative of: (i) (x-a)(x-b) (ii) (ax^2+b)^2 (iii) (x-a)/(x-b).
Hint. Use the product rule for (i) and (ii), and the quotient rule for (iii).
(i) Expand: x^2-(a+b)x+ab. Derivative: 2x-(a+b). (ii) Using product rule with u=v=ax^2+b, u'=v'=2ax: derivative = 2ax(ax^2+b)+(ax^2+b)(2ax) = 4ax(ax^2+b). (iii) Using quotient rule with u=x-a,u'=1,v=x-b,v'=1: derivative = [1.(x-b)-(x-a).1]/(x-b)^2 = (a-b)/(x-b)^2.
✦ Working through each part gives: (i) 2x-(a+b) (ii) 4ax(ax^2+b) (iii) (a-b)/(x-b)^2.
Find the derivative of (x^n-a^n)/(x-a) for some constant a.
Hint. Use polynomial division to rewrite the quotient as a sum of n terms first, then differentiate termwise.
Since x^n-a^n=(x-a)(x^(n-1)+x^(n-2)a+...+a^(n-1)), the quotient simplifies to x^(n-1)+x^(n-2)a+x^(n-3)a^2+...+a^(n-1) for x not equal to a. Differentiating termwise using the power rule: (n-1)x^(n-2)+(n-2)a.x^(n-3)+(n-3)a^2x^(n-4)+...+a^(n-2).
✦ Working through each part gives: (n-1)x^(n-2) + (n-2)a.x^(n-3) + ... + 2a^(n-3)x + a^(n-2).
Find the derivative of: (i) 2x-3/4 (ii) (5x^3+3x-1)(x-1) (iii) x^-3(5+3x) (iv) x^5(3-6x^-9) (v) x^-4(3-4x^-5) (vi) 2/(x+1) - x^2/(3x-1).
Hint. Use the power rule directly for (i); use the product rule for (ii)-(v) after treating negative-power terms with the power rule; use the quotient rule termwise for (vi).
(i) Derivative of a linear function: 2. (ii) Product rule with u=5x^3+3x-1,u'=15x^2+3; v=x-1,v'=1: derivative = (15x^2+3)(x-1)+(5x^3+3x-1) = 20x^3-15x^2+6x-4. (iii) Expand first: x^-3(5+3x)=5x^-3+3x^-2, derivative = -15x^-4-6x^-3. (iv) Expand: x^5(3-6x^-9)=3x^5-6x^-4, derivative=15x^4+24x^-5. (v) Expand: x^-4(3-4x^-5)=3x^-4-4x^-9, derivative=-12x^-5+36x^-10. (vi) Quotient rule on each term separately: d/dx[2/(x+1)]=-2/(x+1)^2; d/dx[x^2/(3x-1)] using u=x^2,u'=2x,v=3x-1,v'=3: [2x(3x-1)-3x^2]/(3x-1)^2=(3x^2-2x)/(3x-1)^2. Subtracting: -2/(x+1)^2 - (3x^2-2x)/(3x-1)^2.
✦ Working through each part gives: (i) 2 (ii) 20x^3-15x^2+6x-4 (iii) -15x^-4-6x^-3 (iv) 15x^4+24x^-5 (v) -12x^-5+36x^-10 (vi) -2/(x+1)^2 - (3x^2-2x)/(3x-1)^2.
Find the derivative of cos x from first principle.
Hint. Use the compound-angle identity cos(A)-cos(B) = -2.sin((A+B)/2).sin((A-B)/2) on cos(x+h)-cos(x), then apply the standard trigonometric limit.
f'(x)=lim(h->0)[cos(x+h)-cos(x)]/h. Using the identity with A=x+h, B=x: cos(x+h)-cos(x) = -2.sin(x+h/2).sin(h/2). Dividing by h: -sin(x+h/2).[sin(h/2)/(h/2)]. As h->0, sin(x+h/2)->sin(x) and sin(h/2)/(h/2)->1.
✦ Working through each part gives: -sin x.
Find the derivative of the following functions: (i) sin(x)cos(x) (ii) sec(x) (iii) 5sec(x)+4cos(x) (iv) cosec(x) (v) 3cot(x)+5cosec(x) (vi) 5sin(x)-6cos(x)+7 (vii) 2tan(x)-7sec(x).
Hint. Use the product rule where sin x and cos x are multiplied directly, and the quotient rule (writing each function in terms of sin x, cos x) for sec, cosec, cot, tan.
(i) Product rule (or the identity sin(x)cos(x)=sin(2x)/2): derivative = cos^2(x)-sin^2(x) = cos(2x). (ii) sec(x)=1/cos(x); quotient rule gives sin(x)/cos^2(x) = sec(x)tan(x). (iii) Termwise: 5.sec(x)tan(x) + 4.(-sin(x)) = 5sec(x)tan(x)-4sin(x). (iv) cosec(x)=1/sin(x); quotient rule gives -cos(x)/sin^2(x) = -cosec(x)cot(x). (v) Termwise using d/dx(cot x)=-cosec^2(x) and d/dx(cosec x)=-cosec(x)cot(x): -3cosec^2(x)-5cosec(x)cot(x). (vi) Termwise: 5cos(x)+6sin(x). (vii) Termwise: 2sec^2(x)-7sec(x)tan(x).
✦ Working through each part gives: (i) cos(2x) (ii) sec(x)tan(x) (iii) 5sec(x)tan(x)-4sin(x) (iv) -cosec(x)cot(x) (v) -3cosec^2(x)-5cosec(x)cot(x) (vi) 5cos(x)+6sin(x) (vii) 2sec^2(x)-7sec(x)tan(x).
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh112.pdf) — Exercise 12.1 (32 questions), Exercise 12.2 (11 questions), plus the chapter's Miscellaneous Exercise (30 questions), 73 questions total, the largest solutions file in this project, surpassing Straight Lines and Conic Sections at 70 each. Confirmed against the CBSE curriculum PDF that the Chain Rule is formative-only and completely absent from the book (zero hits for 'chain rule' or 'composite'), yet Miscellaneous Exercise Q12-13 still require differentiating (ax+b)^n type expressions, solved via a first-principles substitution technique built from the chapter's own standard algebraic limit rather than a general chain rule. Confirmed the main syllabus line names limits of exponential and logarithmic functions as summative despite zero coverage anywhere in the book (full-text search for 'exponential'/'logarithm'/'e^x', zero hits) — a reverse gap matching the pattern found in Sequences and Series' infinite-GP-sum; taught in the .md but not fabricated into this solutions file, which stays strictly faithful to the book's own 73 actual questions. All 32 questions in Exercise 12.1 and the first 32 questions of Exercise 12.2/Miscellaneous were cross-checked via 300dpi page renders given how badly piecewise-function and stacked-fraction notation garbles in raw PDF text extraction. Every limit and derivative answer in this file was independently verified with sympy before being written up as a step-by-step solution, catching zero discrepancies against the hand-derived working.. Questions are referenced from the NCERT textbook for identification.
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