Properties of Differentials Here we consider real-valued functions of real variable. ( ) If f is a constant function, then df = . ( ) If f x ( ) = identity function, then df = ( ) If f is differentiable and c ∈ , then d cf cf ′ ( ) If f g are differentiable, then d f g df dg g x dx ′ + ′ ( ) If f g are differentiable, then d fg fdg gdf f x g x f x g x dx ( ( ) ( )) ( ) If f g are differentiable, then d f g gdf fdg g g x f f x g x g / ′ ′ , where g x ( ) ¹ . ( ) If f g are differentiable and h g is defined, then dh g x g x dx ′ ′ ( ( )) ( ) If h x e f x , then dh ′ ( ) If f x ( ) > for all x and g x log( ( )) , then dg f x dx ′ ( ) and and Differentials and Partial Derivatives Example .
Let f g a b :( , ) → be differentiable functions. Show that d fg fdg gdf ) = Let f g a b :( , ) → be differentiable functions and h x f x g x ( ) ( ) . Then h , being a product of differentiable functions, is differentiable on ( , ) a b . So by definition dh h x dx = ′ ( ) .
Now by using product rule we have ′ ′ ′ h x f x g x x g x ( ) ( ) . Thus dh h x dx = ′ ( ) = ( ( ) ( ) ( )) ( ) ( ) f x g x x g x dx f x g x dx x g x dx ′ ′ ′ ′ = f x dg g x df fdg gdf Example . Let g x . Calculate the differential dg .
Note that g is differentiable and ′ g x Thus dg cos ) Example . If the radius of a sphere, with radius cm, has to decrease by . cm, approximately how much will its volume decrease? We know that volume of a sphere is given by V = , where r > is the radius.
So the differential dV r dr = and hence ∆ ≈ V dV = p ( ) ( . - )cm = . )cm = − π cm . Note that we have used dr = ( .
cm, because radius decreases from to . . Again the negative sign in the answer indicates that the volume of the sphere decreases about cm .