Differentiation: More Quotient Rule Practice NCERT CBSE

Deepen your understanding of differentiation by tackling more complex functions using the quotient rule. This video provides step-by-step solutions for examples involving exponential, logarithmic, and power functions. Perfect for calculus students! #calculus #differentiation #quotientrule #derivatives #math #tutorial ncert cbse

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = 1/(eˣ+1)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
We can rewrite y as y = (eˣ + 1)⁻¹. We will use the chain rule: d/dx(f(g(x))) = f'(g(x))g'(x).
Let f(u) = u⁻¹ and g(x) = eˣ + 1.
First, find the derivatives of f(u) and g(x):
f'(u) = d/du (u⁻¹) = -1u⁻² = -1/u²
g'(x) = d/dx (eˣ + 1) = eˣ + 0 = eˣ
Now, apply the chain rule:
dy/dx = (-1/(eˣ + 1)²) * (eˣ)
dy/dx = -eˣ / (eˣ + 1)²
Alternatively, using the quotient rule:
u(x) = 1, v(x) = eˣ + 1
u'(x) = 0, v'(x) = eˣ
dy/dx = [v(x)u'(x) - u(x)v'(x)] / [v(x)]²
dy/dx = [(eˣ + 1)(0) - (1)(eˣ)] / (eˣ + 1)²
dy/dx = [0 - eˣ] / (eˣ + 1)²
dy/dx = -eˣ / (eˣ + 1)²
Final Answer: dy/dx = -eˣ / (eˣ + 1)²

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = x/(log(x))
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
We will use the quotient rule: dy/dx = [v(x)u'(x) - u(x)v'(x)] / [v(x)]², where u(x) = x and v(x) = log(x). (Assuming natural logarithm, log base e)
First, find the derivatives of u(x) and v(x):
u'(x) = d/dx (x) = 1
v'(x) = d/dx (log(x)) = 1/x
Now, apply the quotient rule:
dy/dx = [(log(x))(1) - (x)(1/x)] / (log(x))²
Simplify the numerator:
dy/dx = [log(x) - 1] / (log(x))²
Final Answer: dy/dx = (log(x) - 1) / (log(x))²

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = 2ˣ/(log(x))
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
We will use the quotient rule: dy/dx = [v(x)u'(x) - u(x)v'(x)] / [v(x)]², where u(x) = 2ˣ and v(x) = log(x). (Assuming natural logarithm, log base e)
First, find the derivatives of u(x) and v(x):
u'(x) = d/dx (2ˣ) = 2ˣlog(2) (using the rule d/dx(aˣ) = aˣ log(a))
v'(x) = d/dx (log(x)) = 1/x
Now, apply the quotient rule:
dy/dx = [(log(x))(2ˣlog(2)) - (2ˣ)(1/x)] / (log(x))²
To simplify the numerator, factor out 2ˣ:
dy/dx = [2ˣ * (log(x)log(2) - 1/x)] / (log(x))²
We can also write (log(x)log(2) - 1/x) as (x log(x)log(2) - 1)/x:
dy/dx = [2ˣ * (x log(x)log(2) - 1)/x] / (log(x))²
dy/dx = 2ˣ(x log(x)log(2) - 1) / [x(log(x))²]
Final Answer: dy/dx = 2ˣ(x log(x)log(2) - 1) / [x(log(x))²]

𝐐𝐮𝐞𝐬𝐭𝐢𝐨𝐧: Find dy/dx if y = (2eˣ - 1) / (2eˣ + 1)
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧:
We will use the quotient rule: dy/dx = [v(x)u'(x) - u(x)v'(x)] / [v(x)]², where u(x) = 2eˣ - 1 and v(x) = 2eˣ + 1.
First, find the derivatives of u(x) and v(x):
u'(x) = d/dx (2eˣ - 1) = 2eˣ
v'(x) = d/dx (2eˣ + 1) = 2eˣ
Now, apply the quotient rule:
dy/dx = [(2eˣ + 1)(2eˣ) - (2eˣ - 1)(2eˣ)] / (2eˣ + 1)²
Expand the numerator:
dy/dx = [4e^(2x) + 2eˣ - (4e^(2x) - 2eˣ)] / (2eˣ + 1)²
dy/dx = [4e^(2x) + 2eˣ - 4e^(2x) + 2eˣ] / (2eˣ + 1)²
Simplify the numerator:
dy/dx = 4eˣ / (2eˣ + 1)²
Final Answer: dy/dx = 4eˣ / (2eˣ + 1)²

𝐀𝐛𝐨𝐮𝐭 𝐌𝐚𝐭𝐡𝐬 𝐏𝐥𝐚𝐭𝐭𝐞𝐫
Are you ready to explore the world of mathematics and statistics? Our channel offers a variety of short, solved example videos that cover everything from high school and college-level math to statistics and quantitative aptitude concepts for entrance exams.