Learn how to find the derivatives (dy/dx) of various functions, including those with multiple products and roots. This video demonstrates the application of the product rule, chain rule, and explores how logarithmic differentiation can simplify complex expressions, alongside straightforward power rule applications. Perfect for calculus students! #calculus #differentiation #derivatives #productrule #chainrule #logarithmicdifferentiation #math #tutorial ncert cbse

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = x(2x-3)sqrt(x-2)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
I'll use logarithmic differentiation for this problem since it involves a product of multiple functions.
Let y = x(2x-3)(x-2)^(1/2)
1. Take the natural logarithm of both sides:
ln(y) = ln[x(2x-3)(x-2)^(1/2)]
2. Expand the right side using logarithm properties:
ln(y) = ln(x) + ln(2x-3) + (1/2)ln(x-2)
3. Differentiate both sides with respect to x:
(1/y) * dy/dx = d/dx(ln(x)) + d/dx(ln(2x-3)) + d/dx((1/2)ln(x-2))
(1/y) * dy/dx = 1/x + (1/(2x-3)) * (2) + (1/2) * (1/(x-2)) * (1)
(1/y) * dy/dx = 1/x + 2/(2x-3) + 1/(2(x-2))
4. Combine the terms on the right side by finding a common denominator:
The common denominator is 2x(2x-3)(x-2).
(1/y) * dy/dx = [1 * 2(2x-3)(x-2) + 2 * x * 2(x-2) + 1 * x(2x-3)] / [2x(2x-3)(x-2)]
Expand the numerator:
= [2(2x^2 - 4x - 3x + 6) + 4x(x-2) + x(2x-3)]
= [2(2x^2 - 7x + 6) + 4x^2 - 8x + 2x^2 - 3x]
= [4x^2 - 14x + 12 + 4x^2 - 8x + 2x^2 - 3x]
Combine like terms:
= 10x^2 - 25x + 12
So, (1/y) * dy/dx = (10x^2 - 25x + 12) / [2x(2x-3)(x-2)]
5. Solve for dy/dx by multiplying both sides by y:
dy/dx = y * (10x^2 - 25x + 12) / [2x(2x-3)(x-2)]
Substitute the original expression for y:
dy/dx = [x(2x-3)(x-2)^(1/2)] * (10x^2 - 25x + 12) / [2x(2x-3)(x-2)]
6. Cancel common terms:
The 'x' and '(2x-3)' terms cancel out.
(x-2)^(1/2) divided by (x-2) simplifies to 1/(x-2)^(1/2) or 1/sqrt(x-2).
dy/dx = (10x^2 - 25x + 12) / [2 * (x-2)^(1/2)]
dy/dx = (10x^2 - 25x + 12) / (2sqrt(x-2))
Final Answer: dy/dx = (10x^2 - 25x + 12)/(2sqrt(x-2))

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = (2x-5)^(3/2) - 7/sqrt(x)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
First, I'll rewrite the function using fractional exponents for easier differentiation:
y = (2x-5)^(3/2) - 7x^(-1/2)
Now, I'll differentiate each term with respect to x.
1. Differentiate the first term, (2x-5)^(3/2):
I'll use the chain rule. Let u = 2x-5.
d/dx [(2x-5)^(3/2)] = (3/2)(2x-5)^(3/2 - 1) * d/dx(2x-5)
= (3/2)(2x-5)^(1/2) * (2)
= 3(2x-5)^(1/2)
= 3sqrt(2x-5)
2. Differentiate the second term, -7x^(-1/2):
I'll use the power rule.
d/dx [-7x^(-1/2)] = -7 * (-1/2)x^(-1/2 - 1)
= (7/2)x^(-3/2)
3. Combine the derivatives of both terms:
dy/dx = 3sqrt(2x-5) + (7/2)x^(-3/2)
Final Answer: dy/dx = 3sqrt(2x-5) + 7/2 x^(-3/2)

𝐀𝐛𝐨𝐮𝐭 𝐌𝐚𝐭𝐡𝐬 𝐏𝐥𝐚𝐭𝐭𝐞𝐫
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