Solution 09

 Given: y = sin⁻¹(2x√(1-x²))

Using the chain rule with d/dx[sin⁻¹(u)] = 1/√(1-u²) · du/dx

Let u = 2x√(1-x²)

Step 1: Find du/dx using the product rule. u = 2x · √(1-x²)

du/dx = 2x · d/dx[√(1-x²)] + √(1-x²) · d/dx[2x]

For d/dx[√(1-x²)] = d/dx[(1-x²)^(1/2)] = (1/2)(1-x²)^(-1/2) · (-2x) = -x/√(1-x²)

So: du/dx = 2x · (-x/√(1-x²)) + √(1-x²) · 2 du/dx = -2x²/√(1-x²) + 2√(1-x²) du/dx = (-2x² + 2(1-x²))/√(1-x²) du/dx = (-2x² + 2 - 2x²)/√(1-x²) du/dx = (2 - 4x²)/√(1-x²)

Step 2: Find 1-u². u² = (2x√(1-x²))² = 4x²(1-x²) = 4x² - 4x⁴

1-u² = 1 - (4x² - 4x⁴) = 1 - 4x² + 4x⁴ = (1-2x²)²

Therefore: √(1-u²) = |1-2x²|

Step 3: Apply the chain rule. dy/dx = 1/√(1-u²) · du/dx dy/dx = 1/|1-2x²| · (2-4x²)/√(1-x²) dy/dx = (2-4x²)/(|1-2x²|√(1-x²)) dy/dx = 2(1-2x²)/(|1-2x²|√(1-x²))

Since we typically assume the domain where the derivative exists, and noting that 1-2x² can be positive or negative:

dy/dx = 2/√(1-x²) when 1-2x² > 0 (i.e., when |x| < 1/√2)

dy/dx = -2/√(1-x²) when 1-2x² < 0 (i.e., when |x| > 1/√2)

The most common simplified form is: dy/dx = ±2/√(1-x²)

solution 6 ,7 and 8

 6. If y = ln((1+x)/(1-x)), find dy/dx and simplify

Using the chain rule with ln(u) where u = (1+x)/(1-x):

dy/dx = 1/u · du/dx = (1-x)/(1+x) · du/dx

Find du/dx using the quotient rule: du/dx = [(1-x)(1) - (1+x)(-1)]/(1-x)² du/dx = [1-x + 1+x]/(1-x)² du/dx = 2/(1-x)²

Therefore: dy/dx = (1-x)/(1+x) · 2/(1-x)² dy/dx = 2/[(1+x)(1-x)] dy/dx = 2/(1-x²)

7. A function is given implicitly by x³ + y³ = 3xy. Find dy/dx

Differentiate both sides with respect to x: d/dx(x³ + y³) = d/dx(3xy)

Left side: 3x² + 3y² · dy/dx

Right side (using product rule): 3[x · dy/dx + y · 1] = 3x · dy/dx + 3y

Setting them equal: 3x² + 3y² · dy/dx = 3x · dy/dx + 3y

Collect terms with dy/dx: 3y² · dy/dx - 3x · dy/dx = 3y - 3x dy/dx(3y² - 3x) = 3(y - x) dy/dx = (y - x)/(y² - x)

8. If y = (x² + 1)^(tan⁻¹x), find dy/dx using logarithmic differentiation

Take the natural logarithm of both sides: ln y = ln[(x² + 1)^(tan⁻¹x)] ln y = tan⁻¹x · ln(x² + 1)

Differentiate both sides with respect to x: (1/y) · dy/dx = d/dx[tan⁻¹x · ln(x² + 1)]

Using the product rule on the right side: d/dx[tan⁻¹x · ln(x² + 1)] = tan⁻¹x · d/dx[ln(x² + 1)] + ln(x² + 1) · d/dx[tan⁻¹x]

= tan⁻¹x · (2x)/(x² + 1) + ln(x² + 1) · 1/(1 + x²)

= (2x tan⁻¹x)/(x² + 1) + ln(x² + 1)/(1 + x²)

Therefore: (1/y) · dy/dx = (2x tan⁻¹x + ln(x² + 1))/(x² + 1)

Multiply both sides by y = (x² + 1)^(tan⁻¹x):

dy/dx = (x² + 1)^(tan⁻¹x) · (2x tan⁻¹x + ln(x² + 1))/(x² + 1)

Solution 3, 4 and 5

 3. If y = tan⁻¹((1+x)/(1-x)), find dy/dx

Using the chain rule with the derivative of tan⁻¹(u) = 1/(1+u²) · du/dx

Let u = (1+x)/(1-x)

First, find du/dx using the quotient rule: du/dx = [(1-x)(1) - (1+x)(-1)]/(1-x)² du/dx = [1-x + 1+x]/(1-x)² du/dx = 2/(1-x)²

Now apply the chain rule: dy/dx = 1/(1 + ((1+x)/(1-x))²) · 2/(1-x)²

Simplify the denominator: 1 + ((1+x)/(1-x))² = [(1-x)² + (1+x)²]/(1-x)² = [1-2x+x² + 1+2x+x²]/(1-x)² = (2+2x²)/(1-x)² = 2(1+x²)/(1-x)²

Therefore: dy/dx = 1/(1+x²)

4. If x² + y² = 25, find dy/dx using implicit differentiation

Differentiate both sides with respect to x: d/dx(x² + y²) = d/dx(25) 2x + 2y(dy/dx) = 0

Solve for dy/dx: 2y(dy/dx) = -2x dy/dx = -x/y

5. If y = e^(sin x) · cos x, find dy/dx

Using the product rule: d/dx(uv) = u'v + uv'

Let u = e^(sin x) and v = cos x

Find u': u' = e^(sin x) · d/dx(sin x) = e^(sin x) · cos x

Find v': v' = -sin x

Apply the product rule: dy/dx = u'v + uv' dy/dx = e^(sin x) · cos x · cos x + e^(sin x) · (-sin x) dy/dx = e^(sin x) · cos²x - e^(sin x) · sin x dy/dx = e^(sin x)(cos²x - sin x)

solution 2

 To differentiate y = x^x, I need to use logarithmic differentiation since both the base and exponent contain x.

Given: y = x^x

Step 1: Take the natural logarithm of both sides. ln y = ln(x^x) ln y = x ln x

Step 2: Differentiate both sides with respect to x. d/dx(ln y) = d/dx(x ln x)

Step 3: Apply the chain rule on the left side and product rule on the right side. Left side: d/dx(ln y) = (1/y) · dy/dx

Right side: d/dx(x ln x) = x · d/dx(ln x) + ln x · d/dx(x) = x · (1/x) + ln x · 1 = 1 + ln x

Step 4: Set them equal. (1/y) · dy/dx = 1 + ln x

Step 5: Solve for dy/dx. dy/dx = y(1 + ln x)

Step 6: Substitute back y = x^x. dy/dx = x^x(1 + ln x)

Therefore: dy/dx = x^x(1 + ln x)

Note: This derivative is defined for x > 0, since we need x^x and ln x to be defined.

Solution 1

 To find the derivative of y = ln(sin x), I'll use the chain rule.

Given: y = ln(sin x)

Using the chain rule: dy/dx = d/dx[ln(sin x)]

The derivative of ln(u) is 1/u · du/dx, where u = sin x.

So: dy/dx = 1/(sin x) · d/dx(sin x)

Since d/dx(sin x) = cos x:

dy/dx = 1/(sin x) · cos x = cos x/sin x

Therefore: dy/dx = cot x

We can also write this as cos x/sin x, but cot x is the most simplified form.

Note: This derivative is defined wherever sin x ≠ 0, which means x ≠ nπ where n is any integer.

Solution Q. 10

 

Given the equation $x=e^{{\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)}$, 

we need to find $\left(\frac{dy}{dx}\right)$ at $x=1$.

First, we take the natural logarithm of both sides:

$$\ln x={\tan }^{-1}\left(\frac{y-x^2}{x^2}\right)$$

Let $u=\frac{y-x^2}{x^2}$. Then the equation becomes:

$$\ln x={\tan }^{-1}(u)$$

Taking the tangent of both sides, we get:

$$u=\tan (\ln x)$$

Substituting back $u=\frac{y-x^2}{x^2}$, we have:

$$\frac{y-x^2}{x^2}=\tan (\ln x)$$

Solving for $y$:

$$y=x^2(1+\tan (\ln x))$$

Next, we differentiate $y$ with respect to $x$:

$$\frac{dy}{dx}=\frac{d}{dx}[x^2(1+\tan (\ln x))]$$

Using the product rule, we get:

$$\frac{dy}{dx}=2x(1+\tan (\ln x))+x^2({\sec }^2(\ln x)\cdot \frac{1}{x})$$

Simplifying the second term:

$$\frac{dy}{dx}=2x(1+\tan (\ln x))+x{\sec }^2(\ln x)$$

Evaluating at $x=1$:

• $\ln (1)=0$

• $\tan (0)=0$

• ${\sec }^2(0)=1$

Substituting these values:

$$\frac{dy}{dx}{\mid }_{x=1}=2\cdot 1\cdot (1+0)+1\cdot 1=2+1=3$$

Thus, the value of $\left(\frac{dy}{dx}\right)$ at $x=1$ is $\boxed{{\displaystyle 3}}$.

SImilarly find for other values of x.