Question:

Differentiate the given function w.r.t. x: sin^3x+cos^6x

Answer:

Step 1: Differentiate sin3x with respect to x.

sin3x = 3sin2xcosx

Step 2: Differentiate cos6x with respect to x.

cos6x = -6sin6x

Step 3: Add the two derivatives.

3sin2xcosx - 6sin6x

Question:

Find dy/dx, if y=12(1−cost),x=10(t−sint),−π/2<t<π/2

Answer:

Step 1: Differentiate y with respect to t:

dy/dt = 12sint

Step 2: Substitute x with 10(t−sint) and differentiate it with respect to t:

dx/dt = 10

Step 3: Divide dy/dt by dx/dt to get dy/dx:

dy/dx = (12sint)/10

Question:

Differentiate the given function w.r.t. x: (3x^2−9x+5)^9

Answer:

Step 1: Take the log of the function.

log[(3x^2−9x+5)^9]

Step 2: Take the derivative of the log of the function with respect to x.

d/dx log[(3x^2−9x+5)^9] = 9/[(3x^2−9x+5)] * (2x−9)

Step 3: Use the chain rule to evaluate the derivative of the original function.

d/dx (3x^2−9x+5)^9 = 9(3x^2−9x+5)^8 * (2x−9)

Question:

Differentiate the given function w.r.t. x: cos(acosx+bsinx), for some constant a and b.

Answer:

Step 1: Use the chain rule to differentiate the function.

Step 2: Differentiate cos(acosx+bsinx) w.r.t. x

Step 3: Take the derivative of the inner function, i.e. acosx+bsinx

Step 4: Use the chain rule again to differentiate acosx+bsinx w.r.t. x

Step 5: Differentiate acosx w.r.t. x

Step 6: Differentiate bsinx w.r.t. x

Step 7: Multiply the derivatives of acosx and bsinx with the derivative of cos(acosx+bsinx).

Step 8: Combine the terms to get the final answer.

Answer: -a sin(acosx + bsinx) - b cos(acosx + bsinx)

Question:

Differentiate the given function w.r.t. x: (5x)^3cos2x

Answer:

Step 1: Rewrite the given function in terms of exponents: 5x^3cos2x

Step 2: Use the power rule to differentiate the function: 15x^2(-2sin2x)

Question:

Differentiate the given function w.r.t. x: x^x+x^a+a^x+a^a , for some fixed a>0 and x>0

Answer:

Step 1: Apply the Chain Rule: d/dx (x^x) + d/dx (x^a) + d/dx (a^x) + d/dx (a^a)

Step 2: Differentiate each term: x^x (ln(x) + 1) + x^a (ln(x)) + a^x (ln(a)) + a^a (ln(a))

Question:

Differentiate the given function w.r.t. x: (logx)^logx, x>1 A (logx)^logx[1/x−log(logx)/x] B (logx)^logx[1/x+log(logx)/x]
C −(logx)^logx[1/x−log(logx)/x]
D −(logx)^logx[1/x+log(2logx)/x]

Answer:

Answer: C

Step-by-step solution:

Step 1: Use the chain rule to differentiate the given function.

Step 2: The function (logx)^logx is a combination of two functions: logx and log(logx).

Step 3: Applying the chain rule, the derivative of (logx)^logx is: (logx)^logx[1/x−log(logx)/x]

Step 4: Since the given function is a negative, the derivative is: −(logx)^logx[1/x−log(logx)/x]

Answer: C

Question:

Differentiate the given function w.r.t. x: cot^−1[√1+sinx+√1−sinx/√1+sinx+√1−sinx], 0 < x <π/2

Answer:

1. Using chain rule, d/dx [cot^−1[√1+sinx+√1−sinx/√1+sinx+√1−sinx]] = d/d[√1+sinx+√1−sinx/√1+sinx+√1−sinx] [cot^−1[√1+sinx+√1−sinx/√1+sinx+√1−sinx]]

2. Using Quotient Rule, d/d[√1+sinx+√1−sinx/√1+sinx+√1−sinx] [cot^−1[√1+sinx+√1−sinx/√1+sinx+√1−sinx]] = (d/dx[√1+sinx+√1−sinx] * (√1+sinx+√1−sinx) - (√1+sinx+√1−sinx) * d/dx[√1+sinx+√1−sinx])/(√1+sinx+√1−sinx)^2

3. Simplifying, = (1/2 * (√1+sinx+√1−sinx) * (cosx) - (√1+sinx+√1−sinx) * 1/2 * (√1+sinx+√1−sinx) * (-sinx))/(√1+sinx+√1−sinx)^2

4. Simplifying further, = (cosx - sinx)/(2 * (√1+sinx+√1−sinx)^2)

Question:

Differentiate the given function w.r.t. x:sin^−1(x√x),0≤x≤1

Answer:

Step 1: Differentiate sin^−1(x√x) w.r.t. x

Step 2: Use the chain rule:

d/dx[sin^−1(x√x)] = d/du[sin^−1(u)] * du/dx

Step 3: Differentiate sin^−1(u) w.r.t. u

d/du[sin^−1(u)] = 1/√(1-u^2)

Step 4: Differentiate x√x w.r.t. x

du/dx = √x + x/2√x

Step 5: Substitute the values of d/du[sin^−1(u)] and du/dx in the chain rule equation

d/dx[sin^−1(x√x)] = (1/√(1-x√x^2)) * (√x + x/2√x)

Step 6: Simplify the equation

d/dx[sin^−1(x√x)] = (1/√(1-x√x^2)) * (√x + x/2√x)

d/dx[sin^−1(x√x)] = (1/√(1-x^2)) * (√x + x/2√x)

d/dx[sin^−1(x√x)] = (√x + x/2√x) / √(1-x^2)

Question:

Differentiate the given function w.r.t. x: cos^−1 x/2/√2x+7, −2<x<2

Answer:

Step 1: Rewrite the given function in terms of cos inverse: cos^−1 (x/2/√2x+7)

Step 2: Apply the chain rule to differentiate the function: d/dx[cos^−1 (x/2/√2x+7)] = -1/√[1-(x/2/√2x+7)^2] * (1/2/√2x+7) * (2/√2-2x/2(2x+7)^2)

Step 3: Simplify the expression: d/dx[cos^−1 (x/2/√2x+7)] = -1/(2√2x+7) * (1/√2x+7 - x/2(2x+7)^2)

Question:

Find dy/dx, if y=sin^−1x+sin^−1√1−x^2,−1≤t≤1

Answer:

1. Rewrite y in terms of inverse trigonometric functions: y = sin^−1x + sin^−1√1−x^2

2. Take the derivative of y with respect to x: dy/dx = (1/√1−x^2)·(−x) + (1/√1−x^2)·(−2x)

3. Simplify the expression: dy/dx = (−3x)/√1−x^2

Question:

Differentiate the given function w.r.t. x: (sinx−cosx)^(sinx−cosx), π/4<x<3π/4

Answer:

Step 1: Differentiate the given function w.r.t. x: (sinx−cosx)^(sinx−cosx)

Step 2: Use the chain rule to differentiate the function: d/dx[(sinx−cosx)^(sinx−cosx)] = (sinx−cosx)^(sinx−cosx) * d/dx[sinx−cosx]

Step 3: Differentiate the inside function: d/dx[sinx−cosx] = cosx + sinx

Step 4: Substitute the derivative of the inside function into the chain rule: d/dx[(sinx−cosx)^(sinx−cosx)] = (sinx−cosx)^(sinx−cosx) * (cosx + sinx)

Step 5: Substitute the given range of x values into the function: d/dx[(sinx−cosx)^(sinx−cosx)] = (sin(π/4)−cos(π/4))^(sin(π/4)−cos(π/4)) * (cos(π/4) + sin(π/4)) - (sin(3π/4)−cos(3π/4))^(sin(3π/4)−cos(3π/4)) * (cos(3π/4) + sin(3π/4))

Question:

Differentiate the given function w.r.t. x: x^x^2−3+(x−3)^x^2, for x>3

Answer:

Given, f(x) = x^x^2−3+(x−3)^x^2

Differentiate w.r.t. x,

f’(x) = (x^x^2−3)’ + (x−3)^x^2'

Now,

(x^x^2−3)’ = (x^x^2)‘−3’

= (x^x^2) (ln x) (x^2) − 0

= x^x^2 ln x x^2

Similarly,

(x−3)^x^2)’ = (x−3)^x^2 (ln (x−3)) (x^2)

Now,

f’(x) = x^x^2 ln x x^2 + (x−3)^x^2 (ln (x−3)) (x^2)

= x^x^2 [ln x + (x−3) ln (x−3)] x^2

Therefore, the differentiated form of the given function is,

f’(x) = x^x^2 [ln x + (x−3) ln (x−3)] x^2, for x>3

Question:

If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=a(cosθ+θsinθ) and y=a(sinθ−θcosθ)

Answer:

1. Differentiate the equation x with respect to θ:

dx/dθ = a(−sinθ + cosθ + θcosθ + sinθ)

2. Differentiate the equation y with respect to θ:

dy/dθ = a(cosθ − θsinθ − sinθ − cosθ)

3. Substitute the values of dx/dθ and dy/dθ in the equation:

dy/dx = (dy/dθ)/(dx/dθ)

4. Simplify the equation:

dy/dx = (a(cosθ − θsinθ − sinθ − cosθ))/(a(−sinθ + cosθ + θcosθ + sinθ))

5. Finally, the answer is:

dy/dx = −1