Question
This is called obtaining the derivative of a function from the first principle or “ab initio”.
Question 3:
Show that f(x) = x2/5 is differentiable at x = 0, then find f’(0).
Solution:
To prove that f is differentiable at x = 0, we have to show that Rf’(0) = Lf’(0).
Now, f(0) = 0, then,
Thus, Rf’(0) = Lf’(0) = 0.
Hence, f is differentiable at x = 0 and f’(0) = 0.
Question 4:
Find the derivative of √[(1 – cos 2x)/(1 + cos 2x)]
Solution:
Question 5:
If x3 + y3 = 3axy, find dy/dx.
Solution:
Given, x3 + y3 = 3axy
Differentiating both sides with respect to x, we get:
3x2 + 3y2 (dy/dx) = 3ay + 3ax (dy/dx)
⇒ {3y2 – 3ax} (dy/dx) = 3ay – 3x2
⇒ dy/dx = (3ax – 3x2)/(3y2 – 3ax)
⇒ dy/dx = (ax – x2.)/(y2 – ax).
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Question 6:
Find the derivative of the function (ex cos3 x sin2x), with respect to x.
Solution:
Let y = ex cos3 x sin2x
Taking logarithms on both sides, we get:
log y = log (ex cos3 x sin2x) = x + 3 log (cos x) + 2 log (sin x)
Differentiating both sides, with respect to x, we get:
1/y. dy/dx = 1 – 3sin x/cos x + 2cos x/ sin x
⇒ dy/dx = y.[1 – 3tan x + 2cot x]
⇒ dy/dx = (ex cos3 x sin2x)[1 – 3tan x + 2cot x].
Question 7:
Prove that curves x = y2 and xy = k cut each other at right angles if 8k2 = 1.
Solution:
If both the curves intersect each other at right angles, then their respective tangents at the point of intersection are also perpendicular to each other; that is, the product of their slopes at the point of intersection is –1.
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Now, x = y2
Differentiating both sides, with respect to x, we get:
1 = 2y (dy/dx) dy/dx = 1/2y = m ⇒ 1 (say) …(i)
Again, xy = k
Differentiating both sides, with respect to x, we get: y + x(dy/dx) = 0 dy/dx = –y/x = m ⇒ 2 (say) …(ii)
On solving the equations of the two equations:
y = k1/3 and x = k2/3
Now, m1 × m2 = –1
⇒ 1/2y × ( –y/x) = – 1
⇒ 1/2k1/3 × ( –k1/3 /k2/3) = –1 –
⇒ ½ k –1/3 + ( –1/3) = – 1
⇒ k –2/3 = 2
⇒ 8k2 = 1
Question 8:
The volume of a cube is increasing at the rate of 16 cm3/s. At what rate is its total surface area increasing when the length of an edge is 6 cm?
Solution:
Let V and S be the volume and total surface area of the cube, and x be the side of the cube.
Given, dV/dt = 16 cm3/s
We have to find out dS/dt at x = 6 cm
Now, V = x3
dV/dt = 3x2 (dx/dt)
⇒ 16 = 3x3 (dx/dt)
⇒ dx/dt = 16/3x2
S = 6x2
dS/dt = 12x dx/dt = 12x × 16/3x2
⇒ dS/dt = 64/x
At x = 6 cm
dS/dt = 64/6 = 10⅔ cm2/s. For any Assignment related queries, Call us/WhatsApp:- +1 (315) 557-6473
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Question 9:
Find: ∫ x8/(1 – x3)1/3 dx.
Solution:
Put (1 – x3) = t then – 3x
Now,
x8/(1 – x3)
Question 10:
Find the area between the circle 4x2 + 4y2 = 9 and the parabola y2 = 4x.
Solution:
Now, 4x2 + 4y2 = 9 represents a circle whose centre is at (0,0) and radius is 3/2 and y2 = 4x represents a rightward parabola, whose vertex is at (0, 0).
Now, let us find the points of intersection
4x2 + 16x – 9 = 0 ( y ∵ 2 = 4x)
⇒ x = ½ and –9/2.
If x = ½ then y = ±√2
And if x = –9/2 then y is imaginary
Thus, the points of intersection are (½ , √2) and (½, –√2)
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