Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Differentiation of Composite Functions
Differentiation of Logarithmic Functions
What is an implicit function?
Differentiation of Implicit Functions
A Direct Formula for computing the derivatives of Implicit Functions
Related Resources
In this section, we shall discuss the concepts of composite, logarithmic and implicit functions. First of all we shall study the nature of these functions and then we shall switch to the differentiation of these functions.
If g and h are two functions defined by y = g(u) and u = h(x) respectively then a function defined by y = g [h(x)] or goh(x) is called a composite function or a function of a function.
Hence, if g(x) and h(x) are two differentiable functions, then fog is also differentiable and hence (fog)’(x) = f’(g(x)).g’(x)
This can be better explained as given below:
Let y be a differential function of u and u is a differential function of x, then
dy/dx = dy/(du) × du/(dx)
Let y = g(u) and u = f(x)
Let Δx be increment in x and Δ u and Δy be the Corresponding increments in u and y respectively.
y + Δ y = g(u + Δu) and u + Δu = f(x + Δx)
Δy = g(u + Δu) - g(u) and Δu = f(x + Δx) - f(x)
Δy/(Δu) = (g (u + Δu)/(Δu) and Δu/(Δx) = (f(x + ?x) – f(x))/?x
Δy/(Δx) = Δy/(Δu).Δu/(Δx)
Applying limits
lim?x→0 ?y/?x = lim?u→0 ?y/?u × lim?u→0 ?u/?x
⇒ dy/dx = dy/du × du/dx
= d/du g(u) × d/dx f(x)
If y = t(u) and u = h(x) then dy/dx = dy/du . du/dx
In fact, this can be extended to any number of chains. Hence, if we have y = (A)B, where A is a function of x and B is a constant then we have
dy/dx = B(A)B-1 . d/dx (A).
Find the derivative of composite function y = sin (cot x).
Assume cot x = u so that we have y = sin u, u = cot x
= d/du (sin u ) × d/dx (cot x)
= cos u.(- cosec2x)
= – cos (cot x). cosec2x
In order to find the derivative of:
(a) a function which is the product or quotient of a number of functions like
y = g1(x)g2(x)g3(x)........ or y = (g1(x)g2(x)g3(x)...... )/ (s1(x)s2(x)s3(x).... )
(b) a function of the form [g(x)]s(x) where g and s are both derivable, students are advised to take the logarithm of the function first followed by its differentiation.
This can also be done as y = [g(x)]s(x) = es(x). ln (g(x)) and then differentiate.
Find dy/dx for y = xx.
The given function is y = xx
Taking logarithm of both the sides we have
ln y = x ln x
Now, differentiating both sides we have
d/dx (ln y) = d/dx (x ln x)
⇒ 1/y . dy/dx = x.(1/x) + ln x
= 1 + ln x
Hence, dy/dx = xx (1 + ln x)
If two variables say u and v are associated by a relation of the form g(u, v) = 0 and it is not possible to express one of them in terms of the other i.e. we cannot express u as a function of v in the form v = ∅ (u) then the functions are said to be implicit functions.
Example: Folium is an example of implicit function. Its equation is
Working rule to find dy/dx i.e. implicit derivative are as follows.
Differentiate the given relation between x and y w.r.t. x
Bring all the terms containing dy/dx on left-hand side and remaining terms in right hand side and then find d/dx.
Corresponding to each and every curve that is presented by an implicit equation, there may exist one or more than one explicit functions representing the equation. As a matter of fact, dy/dx at any point on the curve remains the same whether the differentiation is done explicitly or implicitly.
?
Differentiate the function xy = x3 + y3.
Differentiating w.r.t. x we get
dy/dx (xy) = d/dx x3 + d/dx y3
x dy/dx y.1 = 3x2 + 3y2 d/dx
or (x – 3y2) dy/dx = 3x2 - y
dy/dx = (3x2 - y)/(x – 3y2)
Let us consider a function f(x,y) = 0. By taking all the terms of left side and putting it equal to f(x, y) we can directly compute the derivative using the following result:
If x2 + 2xy + y3 = 4, then find dy/dx.
The given function is x2 + 2xy + y3 = 4
Differentiating both sides w.r.t. x,
2x + 2(x dy/dx + y.1) + 3y2 dy/dx = 0
Hence, dy/dx = – 2(x + y)/(2x + 3y2)
The above questions can also be solved using the formula discussed above.
dy/dx = – (diff. of f w.r.t.x keeping y as constant)/(diff. of f w.r.t.y keeping x as constant)
= – (2x + 2y)/(2x + 3y2).
(d2y)/(dx2) can also be written as y" or y2
It is known as second ordered differential co-efficient.
Evaluate limk→0, h→0 (f (a +"k +" h) – f(a + k) – f(a + h) + f(a))/(h k), where f(x) is a twice differential function.
limk→0 1/k [limk→0 ((f (a +"k +" h) – f(a + k))/h – (f(a + h) + f(a))/k)]
= lim(k→0) 1/k [f' (a + k - f'(a))]
= f" (a)
dy/dx = (dy/dt) / (dx/dt)
dy/dx = 1/(dx/dy)
d2y/dx2 ≠ (d2y/dt2) / (d2x/dt2)
d2y/dx2 ≠ 1/ (d2x/dy2)
If y = (sin-1 x)2, then show that (1 – x2) (d2y)/(dx2) – x dy/dx – 2 = 0.
The given function is y = (sin-1 x)2
dy/dx = d/dx (sin-1 x)2
= 2 (sin-1x)2 -1 d/dx (sin-1x)
= (2 sin-1x)/√(1-x2 ) (-1 < x < 1)
√(1 – x(-2)) dy/dx = 2sin-1x
Differentiating again, with respect to x, we get
Q1. While trying to compute the second derivative
(a) d2y/dx2 = (d2y/dt2) / (d2x/dt2)
(b) d2y/dx2 = (dy/dt) / (d2x/dt2)
(c) d2y/dx2 = d/dx (dy/dx)
(d) d2y/dx2 = (d2y/dt2) .(d2t/dx2)
Q2. The composite function (fog)’(x) is
(a) f’(g(x)).g’(x)
(b) f’(g(x)).g’(x)
(c) f’(g(x)).g’(x)
(d) f’(g(x)).g’(x)
Q3. The second derivative can also be written as
(a) d2y/ dx2 or y” or y2.
(b) d2y/ dx2 or y” or y2.
(c) dy/ dx2 or y” or y2.
(d) d2y/ dx2 or y’ or y2.
Q4. dy/dx at any point on the curve
(a) decreases when the differentiation is done explicitly and increases when it is done implicitly.
(b) increases when the differentiation is done explicitly and decreases when it is done implicitly.
(c) changes when the differentiation is done explicitly or implicitly.
(d) remains the same whether the differentiation is done explicitly or implicitly.
Q5. d/dx (f(y)) =
(a) f”(y) dy/dx
(b) f’(y) dy/dx
(c) f(y). f’(y)
(d) f’(y) d2y/dx2
Q1.
Q2.
Q3.
Q4.
Q5.
(c)
(a)
(d)
(b)
Check here for the Complete Syllabus of IIT JEE Mathematics.
Look into the Past Year Papers with Solutions to get a hint of the kind of questions asked in the exam.
Refer theUseful Books of Mathematics here.
To read more, Buy study materials of Methods of Differentiation comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
General Theorems on Differentiation General...
Differentiation of a Function Given in Parametric...
Solved Examples on Differentiation Example 1: i....
Limits using Differentiation L'Hospital's...
Algebraic Operations on Differentiation and...
Differentiation by Abinitio Differentiation by...
Introduction to Differentiation Table of Content...