Linear Approximation and Differential of a function of several variables: Function of Function Rule

**Function of
Function Rule**

Let *F* be a function of two variables *x* , *y*.
Sometimes these variables may be functions of a single variable having same
domain. In this case, the function *F*
ultimately depends only on one variable. So we should be able to treat this *F* as a function of single variable and
study about *dF/dt* . In fact, this is
not a coincidence, it can be proved that

Suppose that *W* ( *x*, *y*)
is a function of two variables *x* , *y* having partial derivatives ∂*W/*∂*x* , ∂*W/*∂*y* . If both the variables *x* , *y* are differentiable functions of a single
variable *t* , then *W* is a differentiable function of *t* and

Let us
consider an example illustrating the above theorem.

** **

**Example 8.18**

Verify
the above theorem for *F* ( *x*, *y*
) =
*x*^{2} −
2 *y*^{2} +
2*xy *and *x *(*t*)* *=* *cos* t *,* y*(*t
*)* *=* *sin* t*,*
t *∈[0, 2*π** *]* *.

**Solution**

Let *F*(*x*,*y*) = *x*^{2}
– 2*y*^{2} + 2*xy* and *x*(*t*) = cos*t*, *y*(*t*) = sin*t*.

Then *F* ( *x*,
*y* ) = cos^{2}*t* − 2 sin^{2}*t*
+
2 cos *t* sin *t* and thus *F* has becomes
a function of one

variable
*t *. So by using chain rule, we see
that

*dF/dt * = 2 cos *t* (− sin *t*) − 4
sin *t* cos *t* + 2(−sin^{2} *t* + cos^{2} *t*)

= −6
cos *t* sin *t* + 2(− sin^{2} *t* + cos^{2} *t*) .

On the
other hand if we calculate

= 2(cos *t*
+
sin *t*)(−sin *t* ) + 2(cos *t* − 2 sin *t* )(cos *t*)

= −6
cos *t* sin *t* + 2(− sin^{2} *t* + cos^{2} *t*)

= *dF* /*dt*

** **

**Example 8.19**

Let *g* ( *x*,
*y* ) = *x*^{2}
−
*yx* + sin(*x* + *y*), *x* (*t*)
=
*e*^{3t} , *y*(*t* ) = *t*^{2}
, *t* ∈ ℝ. Find *dg/dt.*

**Solution**

We shall
follow the tree diagram to calculate.

Also,
some times our *W* ( *x*, *y*)
will be such that *x* =
*x*(*s*
, *t*) , and *y* = *y*(*s* , *t*)
where *s* , *t* ∈ ℝ. Then *W* can be considered as a function that depends on *s* and *t* . If *x* , *y* both have partial derivatives with
respect to *s* , *t* and *W* has partial
derivatives with respect to *x* and *y* , then we can calculate the partial
derivatives of *W* with respect to *s* and *t* using the following theorem.

Suppose that *W* ( *x*, *y*)
is a function of two variables *x* , *y *having partial derivatives ∂*W/*∂*x* , ∂*W/*∂*y* . If both* *variables *x* = *x*(*s,t*) and *y* = *y*(*s,t*), where *s* , *t* ∈ ℝ, have partial* *derivatives
with respect to both *s* and *t*, then

We omit
the proof. The above theorem is very useful. For instance, consider the
situation in which *x *=* r *cos*θ** *, and* y *=* *sin* **θ** *,*
r *≥* *0* *and* **θ** *∈* *ℝ, (change from cartesian co-ordinate
to polar* *co-ordinate system). The
above theorem can be generalized for functions having *n* number of variables.

Let us
consider an example.

** **

**Example 8.20**

Let *g* ( *x*,
*y* ) = 2 *y* + *x* ^{2} , *x* = 2*r* − *s*, *y* = *r*^{2}
+
2*s* , *r*, *s* ∈
ℝ. Find

**Solution**

Here again we shall use the tree diagram to calculate

Tags : Mathematics , 12th Maths : UNIT 8 : Differentials and Partial Derivatives

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12th Maths : UNIT 8 : Differentials and Partial Derivatives : Function of Function Rule | Mathematics

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