05 ନିରନ୍ତରତା ଏବଂ ଭିନ୍ନତା
ବ୍ୟାୟାମ 05
Question:
Find the derivative of the function given by f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8) and hence find f′(1).
Answer:
f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)
f’(x)= (1+x)(2x+2x^3+4x^5+8x^7) + (1+x^2)(1+x^4)(1+x^8) + (1+x)(1+x^2)(4x^3+8x^7) + (1+x)(1+x^2)(1+x^4)(8x^6)
f’(1)= (1+1)(2+2+4+8) + (1+1)(1+1)(1+1) + (1+1)(1+1)(4+8) + (1+1)(1+1)(1+1)(8)
f’(1)= 24 + 8 + 16 + 8
f’(1)= 56
Question:
Differentiate the function w.r.t. x. x^(xcosx)+x^2+1/x^2−1
Answer:
Step 1: Rewrite the function in terms of exponents: f(x) = x^(xcosx + 2) + 1/x^2 - 1
Step 2: Take the derivative of the function w.r.t. x: f’(x) = (xcosx + 2)x^(xcosx + 1) + (-2)/x^3
Question:
Differentiate the function w.r.t. x. (logx)^x+x^(logx)
Answer:
Step 1: Take the natural logarithm of both sides of the equation.
log((logx)^x + x^(logx))
Step 2: Use the logarithmic differentiation rule to differentiate the equation.
(1/x) (x^(logx) * log(logx) + (logx)^x * (1/logx))
Step 3: Simplify the equation.
(1/x) (x^(logx) * log(logx) + (logx)^x * (1/logx)) = (logx)^x * (log(logx) + 1/logx)
Question:
Differentiate (x^2−5x+8)(x^3+7x+9) in three ways mentioned below, (i) By using product rule (ii) By expanding the product to obtain a single polynomial (iii) By logarithmic differentiation Do they all give the same answer?
Answer:
(i) By using product rule:
d/dx[(x^2−5x+8)(x^3+7x+9)] = (x^2−5x+8) d/dx(x^3+7x+9) + (x^3+7x+9) d/dx(x^2−5x+8)
= (x^2−5x+8)(3x^2+7) + (x^3+7x+9)(2x−5)
= 3x^4+14x^3+53x^2−35x+72
(ii) By expanding the product to obtain a single polynomial:
(x^2−5x+8)(x^3+7x+9)
= x^5 +7x^4−35x^3+53x^2−35x+72
(iii) By logarithmic differentiation:
d/dx[(x^2−5x+8)(x^3+7x+9)]
= (x^2−5x+8)(x^3+7x+9) * (d/dx[ln(x^2−5x+8)] + d/dx[ln(x^3+7x+9)])
= (x^2−5x+8)(x^3+7x+9) * (2x−5 + 3x^2+7)
= 3x^4+14x^3+53x^2−35x+72
Yes, all three methods give the same answer.
Question:
Find dy/dx of the function x^y+y^x=1
Answer:
Step 1: Take the natural logarithm of both sides of the equation:
ln(x^y + y^x) = ln(1)
Step 2: Differentiate both sides of the equation with respect to x:
y(ln(x))(x^y)’ + x(ln(y))(y^x)’ = 0
Step 3: Simplify the equation:
y(ln(x))y’x^(y-1) + x(ln(y))x’y^(x-1) = 0
Step 4: Rearrange the equation to solve for dy/dx:
dy/dx = -(y(ln(x))x^(y-1)) / (x(ln(y))y^(x-1))
Question:
Find dy/dx of (cosx)^y=(cosy)^x
Answer:
- Take the natural logarithm of both sides:
ln (cosx)^y = ln (cosy)^x
- Differentiate both sides with respect to x:
y*(cosx)^y*(-sinx) = x*(cosy)^x*(-siny)
- Divide both sides by (cosx)^y:
y*(-sinx) / (cosx)^y = x*(-siny) / (cosy)^x
- Multiply both sides by (cosy)^x:
y*(-sinx) (cosy)^x / (cosx)^y = x*(-siny)
- Divide both sides by (cosy)^x:
y*(-sinx) / (cosx)^y = x*(-siny) / (cosy)^x
- Simplify both sides:
dy/dx = -sinyy / (cosx)^y + sinxx / (cosy)^x
Question:
Find dy/dx of xy=e^x−y
Answer:
Step 1: Rewrite the equation as y = e^x - xy
Step 2: Take the derivative of both sides with respect to x:
dy/dx = e^x - (x*dy/dx + y)
Step 3: Solve for dy/dx:
dy/dx = (e^x - y) / (x)
Question:
Differentiate the function w.r.t. x. (x+1/x)^x+x^(1+1/x)
Answer:
Step 1: Use the chain rule to write the function as a product of two functions: f(x) = (x+1/x)^x * x^(1+1/x)
Step 2: Differentiate the first function w.r.t. x: f’(x) = x^x * ln(x+1/x)
Step 3: Differentiate the second function w.r.t. x: f’(x) = x^(1+1/x) * (1+1/x) * ln(x)
Step 4: Multiply the two derivatives together: f’(x) = x^x * ln(x+1/x) * x^(1+1/x) * (1+1/x) * ln(x)
Step 5: Simplify the expression: f’(x) = x^(x+1+1/x) * ln(x+1/x) * (1+1/x) * ln(x)
Question:
Differentiate the function w.r.t. x. (sinx)^x+sin^−1√x
Answer:
Step 1: Take the natural logarithm of both sides of the equation to make it easier to differentiate: ln((sinx)^x + sin^−1√x)
Step 2: Differentiate both sides of the equation w.r.t. x: (d/dx) ln((sinx)^x + sin^−1√x)
Step 3: Use the chain rule to differentiate the left side of the equation: (d/dx) ln((sinx)^x + sin^−1√x) = (1/((sinx)^x + sin^−1√x)) * (d/dx)((sinx)^x + sin^−1√x)
Step 4: Differentiate the right side of the equation using the product rule: (d/dx)((sinx)^x + sin^−1√x) = (d/dx)(sinx)^x + (d/dx)sin^−1√x
Step 5: Differentiate both terms on the right side of the equation: (d/dx)(sinx)^x = x*(sinx)^(x-1)(d/dx)(sinx) (d/dx)sin^−1√x = (1/(2√x))*(d/dx)(x^2)
Step 6: Substitute the derivatives into the equation: (d/dx) ln((sinx)^x + sin^−1√x) = (1/((sinx)^x + sin^−1√x)) * (x*(sinx)^(x-1)(d/dx)(sinx) + (1/(2√x))*(d/dx)(x^2))
Step 7: Simplify the equation: (d/dx) ln((sinx)^x + sin^−1√x) = (x*(sinx)^(x-1)cosx + (1/(2√x))*2x)/((sinx)^x + sin^−1√x)
Question:
Differentiate the function w.r.t. x. x^x−2^sinx
Answer:
Step 1: Take the natural logarithm of both sides of the equation: ln(x^x−2^sinx)
Step 2: Use the chain rule to differentiate: (x^x)(lnx)−2^sinx(ln2)cosx
Step 3: Simplify the expression: x^xlnx−2^sinxln2cosx
Question:
Differentiate the function w.r.t. x. √[(x−1)(x−2)/(x−3)(x−4)(x−5)]
Answer:
Step 1: Simplify the function as follows: √[(x-1)(x-2)/(x-3)(x-4)(x-5)] = (x-1)(x-2)/√[(x-3)(x-4)(x-5)]
Step 2: Take the derivative of both sides with respect to x: d/dx [(x-1)(x-2)/√[(x-3)(x-4)(x-5)]]
Step 3: Apply the chain rule to the left side: [d/dx (x-1)(x-2)]/√[(x-3)(x-4)(x-5)] + (x-1)(x-2)[d/dx √[(x-3)(x-4)(x-5)]]
Step 4: Take the derivative of the numerator: [2(x-2) - (x-1)(1)]/√[(x-3)(x-4)(x-5)] + (x-1)(x-2)[(-1/2)((x-3)(x-4)(x-5)^(-1/2))(3x-15)]
Step 5: Simplify the expression: [(2x-4) - (x-1)]/√[(x-3)(x-4)(x-5)] + (x-1)(x-2)(3x-15)((x-3)(x-4)(x-5)^(-1/2))
Question:
Differentiate the function w.r.t. x. (xcosx)^x+(xsinx)^1/x
Answer:
Step 1: Take the natural logarithm of both sides. ln[(xcosx)^x+(xsinx)^1/x]
Step 2: Apply the chain rule. x[(xcosx)^x]⋅ln(xcosx) + (1/x)[(xsinx)^1/x]⋅ln(xsinx)
Step 3: Apply the power rule. x[xcosx]^(x-1)⋅(cosx+xsinx) + (1/x)[xsinx]^(-1/x-1)⋅(cosx-sinx)
Step 4: Simplify. xcosx^(x-1)⋅(cosx+xsinx) + (1/x)sinx^(-1/x-1)⋅(cosx-sinx)
Question:
Differentiate the function w.r.t. x. (x+3)^2.(x+4)^3.(x+5)^4
Answer:
Step 1: Take the derivative of the first term, (x + 3)2.
d/dx [ (x + 3)2 ] = 2(x + 3)
Step 2: Take the derivative of the second term, (x + 4)3.
d/dx [ (x + 4)3 ] = 3(x + 4)2
Step 3: Take the derivative of the third term, (x + 5)4.
d/dx [ (x + 5)4 ] = 4(x + 5)3
Step 4: Multiply the derivatives of the three terms together.
2(x + 3) * 3(x + 4)2 * 4(x + 5)3
Step 5: Simplify the expression.
24(x + 3)(x + 4)2(x + 5)3
Question:
Differentiate the function w.r.t. x. (logx)^cosx
Answer:
Step 1: Take the natural log of both sides: ln [(logx)^cosx]
Step 2: Use the Chain Rule to differentiate the expression: cosx * (1/x) * (1/lnx)
Step 3: Simplify the expression: cosx/xlnx
Question:
Find dy/dx of y^x=x^y
Answer:
Step 1: Take the natural log of both sides of the equation: ln(y^x) = ln(x^y)
Step 2: Use the chain rule to differentiate both sides: (1/y)ln(y) dy/dx = (1/x)ln(x) dy/dx
Step 3: Isolate dy/dx on the left side: dy/dx = (x/y)ln(x/y) ln(y)
Question:
Differentiate the function w.r.t. x. x^sinx+(sinx)^cosx
Answer:
-
Take the derivative of the inner function (sinx): d/dx (sinx) = cosx
-
Take the derivative of the outer function (x^sinx): d/dx (x^sinx) = sinx x^(sinx-1)
-
Take the derivative of the second term (sinx)^cosx: d/dx (sinx)^cosx = cosx (sinx)^(cosx-1) d/dx (sinx)
-
Add the derivatives of the two terms: d/dx (x^sinx+(sinx)^cosx) = sinx x^(sinx-1) + cosx (sinx)^(cosx-1) d/dx (sinx)
-
Simplify: d/dx (x^sinx+(sinx)^cosx) = sinx x^(sinx-1) + cosx (sinx)^(cosx-1) cosx
Question:
Differentiate the function w.r.t. x. cosx.cos2x.cos3x
Answer:
Step 1: Take the derivative of the function w.r.t. x.
Step 2: Apply the product rule to differentiate the function.
Step 3: Simplify the expression to obtain the final result.
Answer: -sin(x)cos2x.cos3x - 2cosx.sin2x.cos3x - 3cosx.cos2x.sin3x
JEE ଅଧ୍ୟୟନ ସାମଗ୍ରୀ (ଗଣିତ)
01 ସମ୍ପର୍କ ଏବଂ କାର୍ଯ୍ୟ
02 ଓଲଟା ଟ୍ରାଇଗୋନେଟ୍ରିକ୍ କାର୍ଯ୍ୟଗୁଡ଼ିକ
03 ମ୍ୟାଟ୍ରିକ୍ସ
04 ନିର୍ଣ୍ଣୟକାରୀ
05 ନିରନ୍ତରତା ଏବଂ ଭିନ୍ନତା
- ବ୍ୟାୟାମ 01
- ବ୍ୟାୟାମ 02
- ବ୍ୟାୟାମ 03
- ବ୍ୟାୟାମ 04
- ବ୍ୟାୟାମ 05
- ବ୍ୟାୟାମ 06
- ବ୍ୟାୟାମ 07
- ବ୍ୟାୟାମ 08
- ବିବିଧ ବ୍ୟାୟାମ
06 ଡେରିଭେଟିକ୍ସର ପ୍ରୟୋଗ
07 ଇଣ୍ଟିଗ୍ରାଲ୍
08 ଇଣ୍ଟିଗ୍ରାଲ୍ସର ପ୍ରୟୋଗ
09 ଭେକ୍ଟର୍
10 ତିନୋଟି ଡାଇମେନ୍ସନାଲ୍ ଜ୍ୟାମିତି
11 ରେଖା ପ୍ରୋଗ୍ରାମିଂ
12 ସମ୍ଭାବନା