05 Continuity and Differentiability
Exercise 04
Question:
Differentiate the given function w.r.t. x. log(logx),x>1
Answer:
Step 1: Use the chain rule,
d/dx(log(logx)) = (1/logx) * (1/x)
Step 2: Simplify,
d/dx(log(logx)) = (1/xlogx)
Question:
Differentiate the given function w.r.t. x. cos(logx+e^x),x>0
Answer:
Step 1: Use Chain Rule to differentiate the given function.
Step 2: Differentiate logx with respect to x.
Step 3: Differentiate e^x with respect to x.
Step 4: Multiply the derivatives from Steps 2 and 3 and add them together.
Step 5: Differentiate cos(logx+e^x) with respect to x using the result from Step 4.
Answer: -sin(logx+e^x)*(1/x + e^x)
Question:
Differentiate the given function w.r.t. x. esin^−1x
Answer:
Step 1: Rewrite the given function in terms of y, where y = sin^−1x
Step 2: Differentiate the function w.r.t. y
Step 3: Use the chain rule to differentiate the function w.r.t. x
Differentiate w.r.t. y: e^y
Using the chain rule: e^y * (1/√(1-x^2))
Question:
Differentiate the given function w.r.t. x. e^x^3
Answer:
Step 1: Rewrite the function as (e^x)^3
Step 2: Use the Chain Rule: (e^x)^3 = (3e^x)(e^x)^2
Step 3: Differentiate each term: (3e^x)’ = 3e^x; (e^x)^2 = 2e^x(e^x)'
Step 4: Combine the terms: (3e^x)’ + (e^x)^2 = 3e^x + 2e^x(e^x)'
Step 5: Differentiate the last term: 2e^x(e^x)’ = 2e^xe^x = 2*e^2x
Step 6: Combine the terms: 3e^x + 2e^2x = 3e^x + 2e^2x
Answer: The derivative of e^x^3 w.r.t. x is 3e^x + 2e^2x.
Question:
Differentiate the given function w.r.t. x. y=log(cose^x)
Answer:
Step 1: Take the natural logarithm of both sides of the equation.
log(y) = log(cose^x)
Step 2: Apply the chain rule on the right side of the equation.
log(y) = x*log(cose)
Step 3: Differentiate both sides of the equation with respect to x.
(d/dx)log(y) = (d/dx)(x*log(cose))
Step 4: Simplify the right side of the equation.
(d/dx)log(y) = log(cose) + x*(d/dx)(log(cose))
Step 5: Apply the chain rule on the right side of the equation.
(d/dx)log(y) = log(cose) + x*(1/cose)*(d/dx)(cose)
Step 6: Simplify the right side of the equation.
(d/dx)log(y) = log(cose) - x*(1/cose^2)
Step 7: The final answer is:
(d/dx)log(y) = log(cose) - x*(1/cose^2)
Question:
Differentiate the given function w.r.t. x. y=e^x+e^x^2+…+e^x^5
Answer:
Step 1: Rewrite the given function as a summation of powers of e^x as follows: y= e^x + e^2x + e^3x + e^4x + e^5x
Step 2: Use the power rule to differentiate the function w.r.t. x as follows: dy/dx = (1e^x) + (2e^2x) + (3e^3x) + (4e^4x) + (5*e^5x)
Step 3: Simplify the expression to get the final answer as follows: dy/dx = e^x + 2e^2x + 3e^3x + 4e^4x + 5e^5x
Question:
Differentiate the given function w.r.t. x. y=sin(tan^−1e^−x)
Answer:
Step 1: Use the chain rule to differentiate the function.
Step 2: Differentiate the inner function tan^−1e^−x with respect to x.
Step 3: Differentiate the outer function sin(tan^−1e^−x) with respect to x.
Step 4: Combine the two derivatives to get the final answer.
Answer: -e^−xcos(tan^−1e^−x)
Question:
Differentiate the given function w.r.t. x. cosx/logx,x>0
Answer:
Given: f(x) = cosx/logx, x > 0
Step 1: Rewrite the function as f(x) = (cosx) (1/logx)
Step 2: Apply the product rule of differentiation to f(x):
f’(x) = (cosx) (1/logx)’ + (1/logx) (cosx)'
Step 3: Differentiate the first term (cosx) (1/logx)':
(cosx) (1/logx)’ = (cosx) (-1/logx^2)
Step 4: Differentiate the second term (1/logx) (cosx)':
(1/logx) (cosx)’ = (-1/logx^2) (cosx)'
Step 5: Differentiate the third term (cosx)':
(cosx)’ = -sinx
Step 6: Substitute the derivatives of the three terms into the equation for f’(x):
f’(x) = (cosx) (-1/logx^2) + (-1/logx^2) (-sinx)
Step 7: Simplify the equation:
f’(x) = -(cosx + sinx)/logx^2
Question:
Differentiate the given function w.r.t. x. y=√e^√x,x>0
Answer:
Step 1: Rewrite y in terms of exponential form: y = e^(1/2√x)
Step 2: Take the derivative of both sides with respect to x: dy/dx = (1/2)e^(1/2√x) × (1/2)x^(-1/2)
Step 3: Simplify the expression: dy/dx = (1/4)e^(1/2√x) x^(-1/2)
Question:
Differentiate the given function w.r.t. x. e^x/sinx
Answer:
Step 1: Rewrite the given function as a product of two functions: e^x and 1/sinx
Step 2: Use the product rule to differentiate the given function:
(e^x)(d/dx (1/sinx)) + (1/sinx)(d/dx (e^x))
Step 3: Differentiate the first term:
(e^x)(-cosx/sinx^2)
Step 4: Differentiate the second term:
(1/sinx)(e^x)
Step 5: Combine the two terms:
(e^x)(-cosx/sinx^2) + (1/sinx)(e^x)
Step 6: Simplify the expression:
e^x(-cosx/sinx^2 + 1/sinx)
JEE NCERT Solutions (Mathematics)
01 Relations and Functions
02 Inverse Trigonometric Functions
03 Matrices
04 Determinants
05 Continuity and Differentiability
- Exercise 01
- Exercise 02
- Exercise 03
- Exercise 04
- Exercise 05
- Exercise 06
- Exercise 07
- Exercise 08
- Miscellaneous Exercises
06 Application of Derivatives
07 Integrals
08 Application of Integrals
09 Vectors
10 Three Dimensional Geometry
11 Linear Programming
12 Probability