How to find #(partial^2z)/(partialx^2# given #z(x,y)=e^(-3x)cosy# ?

Find #(partial^2z)/(partialx^2#
given #z(x,y)=e^(-3x)cosy# using partial differentiation?

1 Answer
Apr 3, 2018

#(del^2(z(x,y)))/(delx^2) = 9cos(y)e^(-3x)#

Explanation:

When you compute #(del(z(x,y)))/(delx)#, you treat #y# and any function of #y# as if it were a constant; this allows you to move it outside the derivative as you would any constant, then differentiate the remaining function as you would a full derivative.

#(del(z(x,y)))/(delx) = (del(e^(-3x)cos(y)))/(delx)#

Treat #cos(y)# as if it were a constant:

#(del(z(x,y)))/(delx) = cos(y)(del(e^(-3x)))/(delx)#

Then differentiate the remaining function as if it were a full derivative:

#(del(z(x,y)))/(delx) = cos(y)(d(e^(-3x)))/(dx)#

#(del(z(x,y)))/(delx) = cos(y)(-3e^(-3x))#

We repeat the process for the second derivative:

#(del^2(z(x,y)))/(delx^2) = (del(cos(y)(-3e^(-3x))))/(delx)#

Move the factors #cos(y)# and #-3# outside the derivative:

#(del^2(z(x,y)))/(delx^2) = -3cos(y)(del(e^(-3x)))/(delx)#

Differentiate the remaining functions as if it were a full derivative:

#(del^2(z(x,y)))/(delx^2) = -3cos(y)(d(e^(-3x)))/(dx)#

#(del^2(z(x,y)))/(delx^2) = -3cos(y)(-3e^(-3x))#

Simplify:

#(del^2(z(x,y)))/(delx^2) = 9cos(y)e^(-3x)#