function of a function

FUNCTION OF A FUNCTION

Sin x is a function of x since the value of sin x depends on the value of the angle x . similarly, sin(2x + 5) is a function of the angle (2x + 5) since the value of of sine depends on the value of this angle ..

i.e. (2x + 5) is a function of (2x + 5)

but (2x + 5) is itself a function of x, since its value depends on x

i.e. (2x + 5) is a function of x

sinx(2x + 5) is therefore a function of a function of x and such a expressions are referred to generally as functions of a function

formula dy/dx = dy/du . du/dx

 

examples

1. differentiate with respect to x, y = cos (5x – 4)

 

 

Solution

We shall say let u = 5x – 4 (that’s assigning x to a variable U)

If U = 5x – 4

Then y = cos U (That’s if we substitute 5x -4 as U)

If we differentiate U with respect to X we shall have

.du/dx = 5

and if we differentiate y with respect to U we shall have

.dy/du = -sin U (from trig. Functions, hope you understand)

dy/du = -sin U is same as

dy/du = -sin (5x – 4) since U = 5x – 4 ( hope you understand, if no,use the comment section)

now substituting back to our formula

 

dy/dx = dy/du . du/dx

 

dy/dx =-sin (5x – 4). 5 (. is same as multiplication)

dy/dx = -5sin (5x – 4)

 

Please follow up this example, try to get it right, its very simple

 

2. differentiate with respect to x : f(x) = sin (1 + x² )

solution

y = sin (1 + x² )

let U = 1+ x²

du/dx = 2x (if we differentiate a constant, its zero)

y = sin U

dy/du = cos U

dy/du = cos (1 + x² )

therefore

dy/dx = dy/du . du/dx

 

dy/dx = cos(1 + x² ) . 2x

dy/dx = 2xcos (1 + x² )

 

3. y = (4x – 3) ⁵

solution

y = (4x – 3) ^{5 (}basic standard form is y = x⁵ , dy/dx = 5x⁴

here (4x – 3) replaces the single x

Therefore dy/dx = 5(4x – 3) ⁴ multiply by the derivative of the function of (4x – 3), which is 4

= 5(4x – 3) ⁴ X 4

=20(4x-3) ⁴

lets try another procedure for this same question

y = (4x – 3) ⁵

let u = 4x – 3

y = u⁵

dy/du = 5u⁴

du/dx = 4

therefore dy/dx = 4.5u⁴

since u = 4x – 3

dy/dx = 4. 5(4x – 3) ⁴

dy/dx = 20(4x – 3) ⁴

(this is function of function, hope you understand)

we shall continue with more examples

now lets move to EXPONENTIAL

Remember that among our standard derivatives we have something like this

Y = e^x dy/dx = e^x

That is, when you differentiate an exponential it doesn’t change, it remains the same

Now we shall employ this along as we solve related questions

EXAMPLES

1.    y = e^sin2x

solution

let u = sin 2x

therefore y = e^u

dy/du = e^u (it wont change,)

du/dx = 2os 2x (trig. Function)

dy/dx = 2cos 2x.e^sin2x

 

2.    if y = e^5x determine dy/dx

solution

let u = 5x

y = e^u

dy/du = e^u

du/dx = 5

therefore dy/dx = 5.e^u

dy/dx = 5e^u

but u = 5x

dy/dx = 5e^5x