1 answer

How do you differentiate #cosx^2/y^2-xy=y/x^2#?

Question:

How do you differentiate #cosx^2/y^2-xy=y/x^2#?

Answers

#=\frac{x(-3xy^3-2x^2\sin (x^2)+2\cos (x^2))}{y^2}#

Explanation:

#cosx^2/y^2 - xy = y/x^2#

#y=(cosx^2/y^2 -xy) x^2#

#dy/dx =(cosx^2/y^2 -xy) x^2 #

Applying product rule as: #(f\cdot g)^'=f^'\cdot g+f\cdot g^'#

#f=\frac{\cos (x^2)}{y^2}-xy, g=x^2#

#=\frac{d}{dx}(\frac{\cos (x^2)}{y^2}-xy)x^2+\frac{d}{dx}(x^2)(\frac{\cos \(x^2)}{y^2}-xy)#

#\frac{d}{dx}(\frac{\cos (x^2\)}{y^2}-xy)# = #=-\frac{2x\sin(x^2)}{y^2}-y#

(Applying sum/difference rule as: #(f\pm g)^'=f^'\pm g^'#
#=\frac{d}{dx}(\frac{\cos(x^2)}{y^2})-\frac{d}{dx}(xy)#
Here,
#=\frac{d}{dx}(\frac{\cos(x^2)}{y^2})# = #-\frac{2x\sin(x^2)}{y^2}#
and,
#\frac{d}{dx}(xy)# = #y#)

#=-\frac{2x\sin (x^2)}{y^2}-y#

Again,
#\frac{d}{dx}(x^2)=2x#

Finally,
#=(-\frac{2x\sin(x^2)}{y^2}-y)x^2+2x(\frac{\cos (x^2)}{y^2}-xy)#

Simplifying it,

#=\frac{x(-3xy^3-2x^2\sin (x^2)+2\cos (x^2))}{y^2}#

.

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