Math Examples of Differential Calculus

F ( x ) = 7 / x⁶ = 7x^-6,
Now, we will differentiate it with respect to ‘x’.
On Differentiating, we get-
=> d ( F ( x  ) )/dx = d ( 7 x^-6 )/dx,
=> dy / dx = 7 x^-5 / -5 = -7 / 5x^5.

We will differentiate with respect to ‘u’ as ‘u’ is an independent variable.
Y = tan ^{– 1} (√ ( 1 + u ² – u )),
We will first substitute ‘u =tan v’ as it makes the differentiation process easier.
Substituting u = tan v.
 We get,
y = tan ^{– 1} (√ ( 1 + tan ² v – tan v),
 = tan ^-1 (( 1 + tan ² v ) – tan v ),
as ( 1 + tan ² x = sec x ),
 y = tan ^{– 1} ( sec v – tan v ),
= tan ^-1 ( ( 1 – cos ( П / 2 – v ) / sin ( П / 2 – v ) ),
( as tan x = sin x / cos x ),
 Hence, y = tan ^{– 1} ( tan 1 / 2 ( П / 2 – v ) ),
= П / 4 – v / 2.
 Again substituting the original value of ‘v’.
that is v = tan ^{– 1} u,
y = П / 2 – 1 / 2 tan ^{– 1} u,
Differentiating the equation with respect to u, we have
y ' = d y / d u = - 1 / 2 (1 + u ²).

Here, we have denominator and numerator too.
We will use the quotient rule of differentiation to simplify the expression.
 If, we have,
 f ( x ) = u ( x ) / v ( x ),
Then, according to quotient rule, we have
f ' ( x ) = u ' ( x ) v ( x ) - u ( x ) v ' ( x ) / v ² ( x ),
Applying this formula to differentiate the expression , we have.
 f ' ( x ) = - 1 * ( 6 x + 1 ) - ( 7 – x ) * 6 / ( 6 x + 5 ) ²,
= - 6 x – 1 – 42 + 6 x / ( 6 x + 5 ) ²,
= - 43 / ( 6 x + 5 ) ².
 

Differentiating the function ‘y’ partially with respect to ‘u’, we have
d y / d u = sin v [ u ⁷ ( - sin v ) + 7 u ⁶ cos u ] + v cos u,
= 7 u ⁶ cos u sin v – sin ² v u ⁷ + v cos u.
 again differentiating the equation with respect to ‘v’ , we have
dy / dv = u ⁷ cos u cos v + sin u,
We have taken ‘u’ as a constant as we are differentiating it partially with respect to ‘v’.

The partial differentiation is done taking ‘u’ and ‘v’ as independent variables.
Partially differentiating the equation with respect to the independent variable ‘u’, we have
d y / d u = sin u ( - sin u ) + cos u ( cos u ) + t . v ⁷. cos u,
We have treated ‘v’ as a constant.
 Now, partially differentiating with respect to ‘v’, we will take ‘u’ as constant.
Hence,
d y / d v = 7 t .v ⁶ . sin u,
We have taken ‘u’ as a constant.

Partial differentiation is same as the normal differentiation.
The only difference is that we take the second variable as a constant during partial differentiation.
d y / d v = 4 v ³ u + u . sin u,
We did not differentiate ‘u’ as it is taken as a constant.
Now, consider the partial differentiation taking ‘v’ as a constant.
We have,
d y / d u = v ⁴ + v [ u cos u + sin u ]. 

Differentiation will be easier if we consider the process in parts-
Let v = x ⁷ - - - – - - - - – - - - - - - - - - - - equation 1
 Differentiating , we get,
d v / d x = 7 x ⁶ - – - - - - - - - - - - - - - -equation 2
as d x ⁿ / d x = n x ^{n – 1}
Taking the second part, we have
v = sin t . x ⁴,
‘sin t’ is a constant as ‘t’ is not an independent variable.
d v / d x = 4 x ³ sin t - - - - - - - - - – - - - -equation 3
As d k / d x = k where ‘k’ is a constant
Taking the third part, we have-
v = sin x,
d v / d x = cos x - ------ - - - - - - - - - – - - - - -equation 4
 Substituting in the original expression, we have
y ' = 7 x ⁶ + sin t . x ⁴ + cos x.

The expression can be differentiated parts by parts.
Taking first part, we have
 v = sin u . cos u
 Differentiating the expression with respect to ‘u’ , we get
d v / d u = d ( sin u . cos u ) / d u,
As we know that,
d ( f ( x ) . g ( x ) ) / d x = f ' ( x ) . g ( x ) + f ( x ) . g ' ( x )
 Hence , d v / d u = cos u . d ( sin u ) / d u + sin u .d ( cos u ) / d u
= cos u . cos u + sin u ( - sin u ),
= cos ² u – sin ² u - - - - - - - - - - - - - - - - - – - ( equation 1 )
 Taking second part , we have
v = 5 u ² . sin u,
Differentiating, we have
d v / d u = d ( 5 u ² . sin u ) / d u,
= sin u . d 5 u ² / d u + 5 u ² d sin u / d u,
Using Product rule
d v / d u = 10 u . sin u + 5 u ² cos u - - - - - - - - – - - - - - -( equation 2 )
Substituting in the original expression, we have
y ' = cos ² u - sin ² u + 10 u . sin u + 5 u ² cos u.

‘u’ is the independent variable in the expression.
We will differentiate the above expression with respect to ‘u’ as ‘u’ is the independent variable.
Considering first part,
let v = u ⁵ - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 1 )
Differentiating, we get
 d v / d u = 5 u ⁴ - - - - - - - - - - - - - - - - - - - - - - - -( equation 2 )
As d x ⁿ / d x = n x ^{n – 1}.
 Now , taking the second part , we have
 v = sin l . u ⁴ - - - - – - - – - - - - - - - - - - - -( equation 3 )
 Here , ‘l’ is a constant.
Differentiating , we get
d v / d u = sin l . d ( u ⁴ ) / d u,
We have taken ‘sin l’ outside the differentiation function as it is a constant.
d v / d u = sin l . (4 u³),
as ( d x ⁿ / d x ) = n x ^{n- 1},
Hence,
d v / d u = 4 sin l .u ³ - - - - - - - - - - - - - - – - - - -( equation 4 )
 Taking the third part in the expression , we have
v = sin l
‘sin l’ is a constant, so
d v / d u = d ( sin l ) / d u
d v / d u = sin l - - - - - - - - - - - - - – - - - - - - - - - -( equation 5 )
as d k / d u = k where k is a constant
Putting the values in the expression, we have
 y ' = 5 u ⁴ + 4 sin l . u ³ + sin l.

We will differentiate it by parts.
First we will differentiate ‘x ⁵’.
Let g ( x ) = x ⁵,
g ' ( x ) = 5 x ⁴,
as, f ' ( x ⁿ ) = n x ^{n – 1}.
Now, taking second part ‘x log n’.
Let g ( x ) = x log n,
g ' ( x ) = log n as ‘log n’ is constant and f ' ( K x ) = K
The third part is ‘log x’.
let g ( x ) = log x,
g ' ( x ) = 1 / x,
as f ' ( log x ) = 1 / x.
Now add all three parts, we get-
f ' ( x ) = d (x ⁵ + x log n + log x ) / d x,
= 5 x ⁴ + log n + 1 / x .