# Differential Calculus Questions and Answers – Implicit Partial Differentiation

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Implicit Partial Differentiation”.

1. Which of the following functions represent an implicit function, where f(x) and ‘y’ are dependent variables and x denotes independent variable?
a) f(x) = x2 + y2 + z2
b) f(x) = 3x2 + 4x + 1
c) f(x) = x3 + 2x2 + 6x
d) f(x) = x3 + x2y + 4y2

Explanation: f(x) = x2 + y2 + z2, f(x) = 3x2 + 4x + 1, f(x) = x3 + 2x2 + 6x, equations are explicit functions. In these explicit functions the function f(x) which is dependent variable is on left – hand side and independent variable ‘x’ is on the right – hand side. But, in f(x) = x3 + x2y + 4y2, both dependent variable ‘y’ and independent variable ‘x’ are on the right – hand side, therefore we call f(x) = x3 + x2y + 4y2 as an implicit function. Differentiating implicit functions are more complex than normal differentiation since they have mixed variables (both independent and dependent variables).

2. Which of the following functions represent an explicit function, where ‘y’ denotes dependent variable and ‘x’ denotes independent variable?
a) y = x3 + x2y + 4y2
b) y = 3x2 + 4x + 1
c) y = 4cos (x) * sin (y)
d) y = x2 – 5xy + 3y2

Explanation: y = x3 + x2y + 4y2, y = 4cos (x) sin (y), y = x2 – 5xy + 3y2, equations are implicit functions. Because, both dependent variable ‘y’ and independent variable ‘x’ are on the right – hand side but, in y = 3x2 + 4x + 1, dependent variable ‘y’ is on left – hand side and independent variable ‘x’ is on the right – hand side. Differentiation gets simple when the function is dependent only on one variable, that is normal differentiation. Therefore, normal differentiation is simple than implicit differentiation.

3. Which of the following statements is true according to implicit differentiation?
a) Implicit differentiation doesn’t use chain rule (chain rule: when f2(x) is function then 2f(x) * f’(x) is the chain rule)
b) In implicit differentiation solving for y’ ($$\frac {dy}{dx}$$) means, y is function of both ‘x’ (independent variable) and ‘y’ (dependent variable)
c) Implicit differentiation is used to find derivatives for explicit function
d) Normal differentiation is difficult than implicit differentiation

Explanation: Generally, we use implicit differentiation when dependent variable and independent variable are on the same side (i.e) the function consists of both dependent and independent variables. Therefore, the statement: In implicit differentiation solving for y’ ($$\frac {dy}{dx}$$) means, y is function of both ‘x’ (independent variable) and ‘y’ (dependent variable) is true.
The statement: Implicit differentiation doesn’t use chain rule (chain rule: when f2(x) is function then 2f(x) * f’(x) is the chain rule) is false because, implicit function always have 2 combined variables where we use chain rule to solve the function.
The statement: Implicit differentiation is used to find derivatives for explicit function is false because, Implicit differentiation is always used to find derivatives for implicit functions only.
The statement: Normal differentiation is difficult than implicit differentiation is false because, when the function contains only one variable, we use normal differentiation, which is simple. But implicit differentiation contains 2 variables to differentiate, which is a bit harder than normal differentiation.

4. Differentiate each function with respect to ‘x’ and solve for y’ that is ($$\frac {dy}{dx}$$). Where x2 + y2 = 64 and x2 – 5xy + 3y2 = 17?
a) y’ = $$\frac {y}{x}$$;         y’ = $$\frac {2x}{7y}$$
b) y’ = –$$\frac {y}{x}$$;         y’ = $$\frac {5x – 2y}{3x}$$
c) y’ = $$\frac {x}{y}$$;         y’ = $$\frac {x}{5x + 6y}$$
d) y’ = –$$\frac {x}{y}$$;         y’ = $$\frac {2x – 5y}{5x – 6y}$$

Explanation: Solving for y’ that is $$\frac {dy}{dx}$$:
Consider x2 + y2 = 64
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$(x2 + y2) = $$\frac {d}{dx}$$(64)
$$\frac {d}{dx}$$(x2) + $$\frac {d}{dx}$$(y2) = 0         {differentiation of a constant is 0}
2x + 2yy’ = 0         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
2(x + yy’) = 0
(x + yy’) = 0
yy’ = -x
y’ = –$$\frac {x}{y}$$
Now, consider x2 – 5xy + 3y2 = 17
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$(x2 – 5xy + 3y2) = $$\frac {d}{dx}$$(17)
$$\frac {d}{dx}$$(x2) – 5$$\frac {d}{dx}$$(xy) + 3$$\frac {d}{dx}$$(y2) = 0         {differentiation of a constant is 0}
2x – 5(xy’ + y) + 3(2yy’) = 0         {‘uv’ differentiation rule: u’v + uv’}
2x – 5xy’ – 5y + 6yy’ = 0
2x – 5y – y’(5x – 6y) = 0
-y’(5x – 6y) = 2x – 5y
y’ = $$\frac {2x – 5y}{(5x – 6y)}$$

5. Find the value of $$\frac {dy}{dx}$$ (or) y’ for the following equation?
Equation: y = 4cos(x) * sin(y) = 1
a) y’ = tan(x)tan(y)
b) y’ = $$\frac {cos⁡(x)cos⁡(y)}{sin⁡(x)sin⁡(y)}$$
c) y’ = $$\frac {1}{sin⁡(x)sin⁡(y)cos⁡(x)cos⁡(y)}$$
d) y’ = 0

Explanation: Solving for y’ that is $$\frac {dy}{dx}$$:
Consider, 4cos(x) * sin(y) = 1
cos(x) * sin(y) = $$\frac {1}{4}$$
$$\frac {d}{dx}$$ (cos(x) * sin(y)) = $$\frac {d}{dx} (\frac {1}{4})$$
cos(x)cos(y’)(y’) + (-sin(x)) sin(y) = 0         {$$\frac {d}{dx}$$ sin(x) = cos(x)}; {$$\frac {d}{dx}$$ cos(x) = -sin(x)}
y’ cos(x)cos(y) = sin(x)sin(y)
y’ = $$\frac {sin⁡(x)sin⁡(y)}{cos⁡(x)cos⁡(y)}$$ {$$\frac {sin}{cos}$$ = tan}
y’ = tan(x)tan(y)
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6. Differentiate the function sin(y) with respect to x and also find the function (5y3 + 2y2 + 2) with respect to x?
a) $$\frac {d}{dx}$$sin(y) = y’ cos(x) $$\frac {d}{dx}$$ (5y3 + 2y2 + 2) = 15y2x’ + 2yx’
b) $$\frac {d}{dx}$$sin(y) = x’ cos(y) $$\frac {d}{dx}$$ (5y3 + 2y2 + 2) = 15x2y’ + 2y’ + 0
c) $$\frac {d}{dx}$$ sin(y) = y’ cos(y) $$\frac {d}{dx}$$ (5y3 + 2y2 + 2) = 15y2y’ + 2y’
d) $$\frac {d}{dx}$$sin(y) = x’ cos(x) $$\frac {d}{dx}$$ (5y3 + 2y2 + 2) = 15x2x’ + 2xx’

Explanation: Solving for $$\frac {d}{dx}$$sin(y):
Consider, sin(y)
Differentiate with respect to ‘x’         {$$\frac {d}{dx}$$ sin(x) = cos(x)}
= Cos(y) * y’         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
$$\frac {d}{dx}$$sin(y) = Cos(y) * y’
Now consider, (5y3 + 2y2 + 2)
Differentiate with respect to ‘x’
$$\frac {d}{dx}$$ (5y3) + $$\frac {d}{dx}$$ (2y2) + $$\frac {d}{dx}$$ (2)
= 15y2y’ + 2y’ + 0         {differentiation of a constant is 0}
= 15y2y’ + 2y’         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
$$\frac {d}{dx}$$ (5y3 + 2y2 + 2) = 15y2y’ + 2y’

7. If x2$$\sqrt {1 + y}$$ + y2$$\sqrt {1 + x}$$ = 0, then find the value of y’ (or) $$\frac {dy}{dx}$$?
a) y’ = $$\frac {4x^3 – y^4 + 4yx^3}{4xy^3 – x^4 + 4y^{3}}$$
b) y’ = $$\frac {4xy^3 – x^4 + 4y^3}{4x^3 – y^4 + 4x^{3}}$$
c) y’ = $$\frac {4xy^3 + x^4 – 4y^3}{4x^3 + y^4 – 4x^{3}}$$
d) y’ = $$\frac {4x^3 + y^4 – 4x^3}{4xy^3 + x^4 – 4y^{3}}$$

Explanation: Consider, x2$$\sqrt {1 + y}$$ + y2$$\sqrt {1 + x}$$ = 0
x2$$\sqrt {1 + y}$$ = -(y2$$\sqrt {1 + x}$$)
Squaring on both sides
x4(1 + y) = y4(1 + x)
x4 + x4y = y4 + xy4
x4 – y4 = xy4 – x4y
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$ (x4 – y4) = $$\frac {d}{dx}$$ (xy4 – x4y)
$$\frac {d}{dx}$$ (x4) – $$\frac {d}{dx}$$ (y4) = $$\frac {d}{dx}$$ (xy4) – $$\frac {d}{dx}$$ (x4y)
4x3 – 4y3y’ = x(4y3y’) + y4 – (x4y’ + y(4x3))
4x3 – 4y3y’ = 4xy3y’ + y4 – x4y’ – 4x3y
4xy3y’ – x4y’ + 4y’y3 = 4x3 – y4 + 4x3y
y’(4xy3 – x4 + 4y3) = 4x3 – y4 + 4x3y
y’ = $$\frac {4x^3 – y^4 + 4yx^3}{4xy^3 – x^4 + 4y^{3}}$$

8. If x4 + y4 = 4axy, then find the value of y’ (or) $$\frac {dy}{dx}$$?
a) y’ = $$\frac {ay – x^3}{y^3 – ax}$$
b) y’ = $$\frac {ax – y^3}{y^3 – ay}$$
c) y’ = $$\frac {ay – y^3}{x^3 – ay}$$
d) y’ = 0

Explanation: Consider, x4 + y4 = 4axy
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$ (x4 + y4) = $$\frac {d}{dx}$$ (4axy)
4x3 + 4y3y’ = 4a(xy’ + y)         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
4y3y’ – 4axy’ = 4ay – 4x3
y’(4y3 – 4ax) = 4ay – 4x3
y’ = $$\frac {4ay – 4x^{3}}{(4y^3 – 4ax)}$$
y’ = $$\frac {ay – x^3}{y^3 – ax}$$

9. Differentiate the function cot(xy) = $$\frac {3}{4}$$ with respect to ‘x’?
a) y’ = $$\frac {y}{x}$$;
b) y’ = –$$\frac {y}{x}$$
c) y’ = $$\frac {x}{y}$$;
d) y’ = –$$\frac {x}{y}$$

Explanation: Consider, cot(xy) = $$\frac {3}{4}$$
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$ cot(xy) = $$\frac {d}{dx} (\frac {3}{4})$$
(-cosec2(xy)) * (xy’ + y) = 0         {differentiation of a constant is 0}
-xy’ cosec2 (xy) – ycosec2(xy) = 0         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
xy’ cosec2 (xy) + ycosec2(xy) = 0 {$$\frac {d}{dx}$$ cot(x) = – cosec2(x)}
y’ = $$\frac {- y cosec^2 (xy)}{xcosec^2 (xy)}$$
y’ = –$$\frac {y}{x}$$

10. Differentiate the functions (x + y) = sin(xy) with respect to ‘x’?
a) y’ = $$\frac {ycos(xy) – 1}{1 – xcos(xy)}$$
b) y’ = $$\frac {1 + ycos(xy)}{2 + xcos(xy)}$$
c) y’ = $$\frac {ysin(x)}{xcos(xy)}$$
d) y’ = $$\frac {ysin(xy) – 1}{2sin⁡(xy)}$$

Explanation: Consider, (x + y) = sin(xy)
Differentiate w. r. t ‘x’ on both sides
$$\frac {d}{dx}$$ (x + y) = $$\frac {d}{dx}$$ sin(xy)
1 + y’ = xcos(xy)y’ + ycos(xy)         {Chain rule for any function [f(g(x))]’ = f’(g(x)) * g’(x)}
y’ (1 – x cos(xy)) = y cos(xy) – 1 {$$\frac {d}{dx}$$ sin(x) = cos(x)}
y’ = $$\frac {ycos(xy) – 1}{1 – xcos(xy)}$$

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