Differential calculus questions with solutions are provided for students to practise differentiation questions. Differential calculus is a branch of Calculus in mathematics that studies the instantaneous rate of change in a function corresponding to a given input value. Geometrically, it represents the slope of the tangent line to the graph of the function at a given particular point, provided that the function exists and is differentiable at that point. The derivative of a real-valued function at a given point in its domain represents the closest linear approximation of that function at that point.
Learn more about differential calculus in maths.
In the above figure, y = f(x) is a continuous and differentiable function between x and x + h, then derivative of y, dy/dx is the slope of the tangent to the graph of y at x.
Then, the derivative of a function is defined as:
\(\begin{array}{l}\frac{d}{dx}f(x)=f'(x)=\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\end{array} \) |
Derivatives of Some Functions in Differential Calculus
| Function f(x) |
Derivative f’(x) |
Function f(x) |
Derivative f’(x) |
| xn |
nxn – 1 |
sec x |
sec x.tan x |
| axn; a is a constant |
a.n.xn – 1 |
cosec x |
–cosec x cot x |
| sin x |
cos x |
cot x |
–cosec2 x |
| cos x |
–sin x |
loge x |
x–1 |
| tan x |
sec2 x |
ax |
ax loge a |
| ex |
ex |
loga x |
(x loge a)–1 |
| sin-1 x |
[√(1 – x2)]–1 |
cos–1 x |
–[√(1 – x2)]–1 |
| tan–1 x |
1/(1 + x2) |
cot–1 x |
–1/(1 + x2) |
| sec–1 x |
1/[x √(x2 – 1)] |
cosec–1 x |
–1/[x √(x2 – 1)] |
Some Rules of Differential Calculus
Let us study some essential rules that we are going to require while differentiating any function.
- Differential of addition or subtraction of functions
\(\begin{array}{l}\frac{d}{dx}(f(x) \pm g(x))=\frac{d}{dx}f(x)\pm \frac{d}{dx}g(x)\end{array} \)
- Differential of product of functions
\(\begin{array}{l}\frac{d}{dx}(f(x) . g(x))=\left [ \frac{d}{dx}f(x) \right ]g(x)+ f(x)\left [ \frac{d}{dx}g(x) \right ]\end{array} \)
- Quotient rule of differentiation
\(\begin{array}{l}\frac{d}{dx}\left [ \frac{f(x)}{g(x)} \right ]=\frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{[g(x)]^{2}}\end{array} \)
- Chain rule of differentiation
Let y = f(u) and u = g(x), then
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{du}f(u)\frac{d}{dx}g(x)\end{array} \)
Watch the Video on Theorems of Differentiation
Differential Calculus Questions with Solutions
Solve the following differential calculus question and check your solution with the one given here. Practising these questions will improve your understanding of differentiation and help you score better in examinations. These questions are provided keeping in view the syllabus of Classes XI and XII.
Question 1:
Differentiate the following functions with respect to x:
(i) sin 4x
(ii) cos x3
(iii) x3 + tan x
Solution:
(i) Let f(x) = sin 4x, put 4x = t, then
\(\begin{array}{l}\frac{d}{dx}f(x)=\frac{d}{dt}sin\;t\frac{dt}{dx}\end{array} \)
\(\begin{array}{l}=cos\;4x\frac{d}{dx}4x=4.cos\;4x\end{array} \)
\(\begin{array}{l}\therefore \frac{d}{dx}sin\;4x=4.cos\;4x\end{array} \)
(ii) Let f(x) = cos x3, put x3 = u, then
\(\begin{array}{l}\frac{d}{dx}f(x)=\frac{d}{du}cos\;u\frac{du}{dx}\end{array} \)
\(\begin{array}{l}=-sin\;x^{3}.\frac{d}{dx}x^{3}\end{array} \)
\(\begin{array}{l}=-sin\;x^{3}.3x^{2}\end{array} \)
\(\begin{array}{l}\therefore cos\;x^{3}=-3x^{2}sin\;x^{3} \end{array} \)
(iii) Let h(x) = x3 + tan x, then
\(\begin{array}{l}\frac{d}{dx}h(x)= \frac{d}{dx}\left ( x^{3}+tan\;x \right )=\frac{d}{dx}x^{3}+\frac{d}{dx}tan\;x\end{array} \)
\(\begin{array}{l}=3x^{2}+sec^{2}\;x\end{array} \)
\(\begin{array}{l}\therefore \frac{d}{dx}(x^{3}+tan\;x)=3x^{2}+sec^{2}\;x\end{array} \)
Question 2:
If f(x) = [cos x –sin x]/[cos x + sin x], then prove that f’(x) + [f(x)]2 = –1.
Solution:
Given f(x) = [cos x –sin x]/[cos x + sin x]
Divide both the denominator and numerator by cos x, and we get
f(x) = [1 – tan x]/[1 + tan x] = tan ( 𝜋/4 – x)
Now,
\(\begin{array}{l}f'(x)=\frac{d}{dx}f(x)= \frac{d}{dx}tan\left ( \frac{\pi}{4}-x \right )=-sec^{2}\left ( \frac{\pi}{4}-x \right )\end{array} \)
And [f(x)]2 = tan2( 𝜋/4 – x)
f’(x) + [f(x)]2 = –sec2( 𝜋/4 – x) + tan2( 𝜋/4 – x) = –sec2( 𝜋/4 – x) + sec2( 𝜋/4 – x) –1 = –1
∴ f’(x) + [f(x)]2 = –1
Question 3:
Differentiate with respect to x: (2x + 1)/(2x + 3)
Solution:
Let y = (2x + 1)/(2x + 3) = 1 – 2/(2x + 3)
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left [ 1-\frac{2}{2x+3} \right ]=-2.(-1)(2x+3)^{-2}.2=\frac{4}{(2x+3)^{2}}\end{array} \)
| Second Order Derivative:
Second order derivative of a function means successively differentiating the given function two times. It is denoted as d2y/dx2 where
\(\begin{array}{l}\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left ( \frac{dy}{dx} \right )\end{array} \)
Geometrically, the second order derivative of a function corresponds to curvature or concavity of the graph of the function. This feature is used to determine the maxima, minima and saddle points of a function.
Learn more about second-order derivatives. |
Question 4:
If y = tan–1 [(4x)/(1 + 5x2)] + tan–1[(2 + 3x)/(3 – 2x)], find dy/dx.
Solution:
Given,
\(\begin{array}{l}y=tan^{-1}\left ( \frac{4x}{1+5x^{2}} \right )+tan^{-1}\left ( \frac{2+3x}{3-2x} \right )\end{array} \)
\(\begin{array}{l}=tan^{-1}\left ( \frac{5x-x}{1+5x.x} \right )+tan^{-1}\left ( \frac{\frac{2}{3}+x}{1-\frac{2}{3}x} \right )\end{array} \)
\(\begin{array}{l}=tan^{-1}\left ( \frac{5x-x}{1+5x.x} \right )+tan^{-1}\left ( \frac{\frac{2}{3}+x}{1-\frac{2}{3}.x} \right )\end{array} \)
\(\begin{array}{l}\Rightarrow y=\left [ tan^{-1}5x-tan^{-1}x \right ]+\left [ tan^{-1}\frac{2}{3}+tan^{-1}x \right ]=tan^{-1}5x+tan^{-1}\frac{2}{3}\end{array} \)
Differentiating both sides with respect to x, we get
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left [ tan^{-1}5x+tan^{-1}\frac{2}{3} \right ]= \frac{1}{1+25x^{2}}.5+0=\frac{5}{1+25x^{2}}\end{array} \)
Question 5:
Differentiate the following function:
\(\begin{array}{l}\frac{3x+2}{(x+5)(2x+1)+3}\end{array} \)
Solution:
\(\begin{array}{l}Let\:y=\frac{3x+2}{(x+5)(2x+1)+3}=\frac{3x+2}{2x^{2}+11x+15+3}=\frac{3x+2}{2x^{2}+11x+8}\end{array} \)
Differentiating with respect to x, we get
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left [ \frac{3x+2}{2x^{2}+11x+8} \right ]=\frac{(2x^{2}+11x+8)\frac{d}{dx}(3x+2)-(3x+2)\frac{d}{dx}(2x^{2}+11x+8)}{(2x^{2}+11x+8)^{2}}\end{array} \)
\(\begin{array}{l}= \frac{(2x^{2}+11x+8)(3.1+2)-(3x+2)(4x+11)}{(2x^{2}+11x+8)^{2}}\end{array} \)
\(\begin{array}{l}=\frac{6x^{2}+33x+24-(12x^{2}+41x+22)}{(2x^{2}+11x+8)^{2}}=\frac{-6x^{2}-8x+2}{(2x^{2}+11x+8)^{2}}\end{array} \)
\(\begin{array}{l}\therefore \frac{dy}{dx}=-\frac{2(3x^{2}+4x-1)}{(2x^{2}+11x+8)^{2}}\end{array} \)
Also check:
Question 6:
Find the differential of the following exponential and logarithmic functions:
(i) 5x – 3 cos x + log x
(ii) ex sec x
(iii) 3x/(2 + sin x)
Solution:
(i) Let y = 5x – 3 cos x + log x
Differentiating both sides with respect to x, we get
dy/dx = 5x log 5 + 3 sin x + 1/x.
(ii) Let y = ex sec x
Differentiating both sides with respect to x, we get
dy/dx = d/dx [ex sec x] = sin x {d/dx (ex} + ex {d/dx (sec x)}
= ex sec x + ex.sec x tan x
= ex sec x (tan x + 1)
(iii) Let y = 3x/(2 + sin x)
Differentiating both sides with respect to x, we get
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left [ \frac{3^{x}}{2+sin\;x} \right ]=\frac{(2+sin\;x)\frac{d}{dx}3^{x}-3^{x}\frac{d}{dx}(2+sin\;x)}{(2+sin\;x)^{2}}\end{array} \)
\(\begin{array}{l}=\frac{(2+sin\;x).3^{x}log\;3-3^{x}(0+cos\;x)}{(2+sin\;x)^{2}}\end{array} \)
\(\begin{array}{l}\therefore \frac{dy}{dx}=\frac{3^{x}[(2+sin\;x)log\;3-cos\;x]}{(2+sin\;x)^{2}}\end{array} \)
Question 7:
Differentiate x3/(1 – x3) with respect to x3.
Solution:
Let y = x3/(1 – x3) and put t = x3, thus we have to find the value of dy/dt
\(\begin{array}{l}Then,\:\:\frac{dy}{dt}=\frac{dy}{dx}.\frac{dx}{dt}\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left [ \frac{x^{3}} {1-x^{3}}\right ]=\frac{(1-x^{3})\frac{d}{dx}x^{3}-x^{3}\frac{d}{dx}(1-x^{3})}{(1-x^{3})^{2}}\end{array} \)
\(\begin{array}{l}= \frac{(1-x^{3}).3x^{2}-x^{3}.(0-3x^{2})}{(1-x^{3})^{2}}=\frac{3x^{2}}{(1-x^{3})^{2}}\end{array} \)
And dt/dx = 3x2
\(\begin{array}{l}\therefore \frac{dy}{dt}=\frac{3x^{2}}{(1-x^{3})^{2}}\times \frac{1}{3x^{2}}=\frac{1}{(1-x^{3})^{2}};where\;x\neq 1,0.\end{array} \)
Question 8:
Differentiate with respect to x:
\(\begin{array}{l}(sin\;x)^{sin^{-1}\;x}\end{array} \)
Solution:
\(\begin{array}{l}Let\;y=(sin\;x)^{sin^{-1}\;x}\end{array} \)
Taking logarithm on both sides, we get
\(\begin{array}{l}log\;y={sin^{-1}\;x}\;log\;(sin\;x)\end{array} \)
Differentiating both sides with respect to x, we get
\(\begin{array}{l}\frac{1}{y}.\frac{dy}{dx}=sin^{-1}\;x.\frac{1}{sin\;x}.cos\;x+log(sin\;x).\frac{1}{\sqrt{1-x^{2}}}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=y\left ( sin^{-1}\;x.\frac{1}{sin\;x}.cos\;x+log(sin\;x).\frac{1}{\sqrt{1-x^{2}}} \right )\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=(sin\;x)^{sin^{-1}\;x}\left ( sin^{-1}\;x.\frac{1}{sin\;x}.cos\;x+log(sin\;x).\frac{1}{\sqrt{1-x^{2}}} \right )\end{array} \)
Check out: Derivative calculator
Question 9:
Given function
\(\begin{array}{l}y=x^{y^{x}}\end{array} \)
Prove that differential of y with respect to x is
\(\begin{array}{l}\frac{y\;log\;y(x\;log\;x\;log\;y+1)}{(1-x\;log\;y)x\;log\;x}\end{array} \)
Solution:
Given
\(\begin{array}{l}y=x^{y^{x}}\end{array} \)
Taking logarithms on both sides, we get
log y = yx log x, again taking logarithm on both sides, we have
log (log y) = x log y + log (log x)
Let us differentiate both sides with respect x; we get
\(\begin{array}{l}\frac{1}{log\;y}.\frac{1}{y}=x.\frac{1}{y}\frac{dy}{dx}+log\;y.1+\frac{1}{log\;x}.\frac{1}{x}\end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{1}{y\;log\;y}-\frac{x}{y} \right )\frac{dy}{dx}=\frac{x\;log\;x\;log\;y+1}{x\;log\;x}\end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{1-x\;log\;y}{y\;log\;y} \right )\frac{dy}{dx}=\frac{x\;log\;x\;log\;y+1}{x\;log\;x}\end{array} \)
\(\begin{array}{l}\Rightarrow \frac{dy}{dx}=\frac{y\;log\;y\left ( x\;log\;x\;log\;y+1 \right )}{x\;log\;x(1-x\;log\;y)}\end{array} \)
Question 10:
If x = sin u and y = sin bu, where b is any real constant. Prove that
\(\begin{array}{l}(1-x^{2})\frac{d^{2}y}{dx^{2}}-x.\frac{dy}{dx}+b^{2}y =0\end{array} \)
Solution:
First, we shall determine dx/du and dy/du. Thus, differentiating both x and y with respect to u, we get
\(\begin{array}{l}\frac{dx}{du}=cos\;u\:\:\:and\:\:\:\frac{dy}{du}=b\;cos\;bu\end{array} \)
Now,
\(\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{du}}{\frac{dx}{du}}=\frac{b\;cos\;bu}{cos\;u}\end{array} \)
On squaring both sides, we get,
\(\begin{array}{l}\left ( \frac{dy}{dx} \right )^{2}=\left ( \frac{b.cos\;bu}{cos\;u} \right )^{2}= \frac{b^{2}\;cos^{2}\;bu}{cos^{2}\;u} \end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{dy}{dx} \right )^{2}= \frac{b^{2}(1-sin^{2}\;bu)}{1-sin^{2}\;u} \end{array} \)
\(\begin{array}{l}\Rightarrow \left ( \frac{dy}{dx} \right )^{2}= \frac{b^{2}(1-y^{2})}{1-x^{2}} \:\:\:\:\:(given)\end{array} \)
\(\begin{array}{l}\Rightarrow (1-x^{2})\left ( \frac{dy}{dx} \right )^{2}= b^{2}(1-y^{2})\:\:\:\:…..(i)\end{array} \)
Differentiating both sides of (i) with respect to x, we get
\(\begin{array}{l} (1-x^{2}).2\left ( \frac{dy}{dx} \right ).\frac{d^{2}y}{dx^{2}}+(0-2x).\left ( \frac{dy}{dx} \right )^{2}= b^{2}(0-2y.\frac{dy}{dx})\end{array} \)
Dividing both sides by 2(dy/dx), we get,
\(\begin{array}{l}(1-x^{2}).\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}=-b^{2}y\end{array} \)
\(\begin{array}{l}\Rightarrow (1-x^{2}).\frac{d^{2}y}{dx^{2}}-x\frac{dy}{dx}+b^{2}y=0\end{array} \)
Recommended Video
Practice Questions on Differential Calculus Questions
1. Find the differential of the following functions with respect to x.
\(\begin{array}{l}(i)\:\:y=sin\left ( 2\;tan^{-1}\sqrt{\frac{1-x}{1+x}} \right )\end{array} \)
\(\begin{array}{l}(ii)\:\:y=tan\;(x\;e^{x^{2}})\end{array} \)
\(\begin{array}{l}(iii)\:\:y=e^{x}\;log\;x\end{array} \)
\(\begin{array}{l}(iii)\:\:y=\sqrt{\frac{sec\;x+1}{sec\;x-1}}\end{array} \)
2. Find the derivative of sin (sin x3) at x = 𝜋/2.
3. Find the second order derivative with respect to x of the function x3 + 24xy + y3 = 8.
4. If y = 𝛼eax + 𝛽e –ax, prove that y” – a2y = 0.
5. Differentiate with respect to x: tan (xx).
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