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Calculus Examples
ey=xyey=xy
Step 1
Differentiate both sides of the equation.
ddx(ey)=ddx(xy)ddx(ey)=ddx(xy)
Step 2
Step 2.1
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=ex and g(x)=y.
Step 2.1.1
To apply the Chain Rule, set u as y.
ddu[eu]ddx[y]
Step 2.1.2
Differentiate using the Exponential Rule which states that ddu[au] is auln(a) where a=e.
euddx[y]
Step 2.1.3
Replace all occurrences of u with y.
eyddx[y]
eyddx[y]
Step 2.2
Rewrite ddx[y] as y′.
eyy′
eyy′
Step 3
Step 3.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x and g(x)=y.
xddx[y]+yddx[x]
Step 3.2
Rewrite ddx[y] as y′.
xy′+yddx[x]
Step 3.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
xy′+y⋅1
Step 3.4
Multiply y by 1.
xy′+y
xy′+y
Step 4
Reform the equation by setting the left side equal to the right side.
eyy′=xy′+y
Step 5
Step 5.1
Reorder factors in eyy′.
y′ey=xy′+y
Step 5.2
Subtract xy′ from both sides of the equation.
y′ey-xy′=y
Step 5.3
Factor y′ out of y′ey-xy′.
Step 5.3.1
Factor y′ out of y′ey.
y′(ey)-xy′=y
Step 5.3.2
Factor y′ out of -xy′.
y′(ey)+y′(-x)=y
Step 5.3.3
Factor y′ out of y′(ey)+y′(-x).
y′(ey-x)=y
y′(ey-x)=y
Step 5.4
Divide each term in y′(ey-x)=y by ey-x and simplify.
Step 5.4.1
Divide each term in y′(ey-x)=y by ey-x.
y′(ey-x)ey-x=yey-x
Step 5.4.2
Simplify the left side.
Step 5.4.2.1
Cancel the common factor of ey-x.
Step 5.4.2.1.1
Cancel the common factor.
y′(ey-x)ey-x=yey-x
Step 5.4.2.1.2
Divide y′ by 1.
y′=yey-x
y′=yey-x
y′=yey-x
y′=yey-x
y′=yey-x
Step 6
Replace y′ with dydx.
dydx=yey-x