If you've been given:\(x\cdot \sqrt{x}\)
You have to first change it into fractional exponents:\(x\cdot x^{\frac{1}{2}}\)
Then use the exponent laws to combine the two factors:\(x^{1+\frac{1}{2}}\)
\(x^{\frac{3}{2}}\)
Apply to the definition of a derivative:\(\lim _{x\rightarrow 0}\ \frac{\left(x+\Delta x\right)^{\frac{3}{2}}-x^{\frac{3}{2}}}{\Delta x}\)
Now the complicated part. We will use this relationship first called a difference of squares:\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
Set a and b:\(a=\left(x+\Delta x\right)^{\frac{3}{2}}\)
Square both sides\(a^2=\left(x+\Delta x\right)^3\)
Now b:\(b=x^{\frac{3}{2}}\)
Square both sides:\(b^2=x^3\)
Back to the difference of squares, rearrange for a minus b:\(a-b=\frac{a^2-b^2}{a+b}\)
Substitute what we set a and b into the equation above:\(\frac{a^2-b^2}{a+b}\rightarrow \ \frac{\left[\left(x+\Delta x\right)^{\frac{3}{2}}\right]^2-\left[x^{\frac{3}{2}}\right]^2}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]}\)
Now apply this into the top part of the limit!\(\lim _{x\rightarrow 0}\ \frac{\frac{\left[\left(x+\Delta x\right)^{\frac{3}{2}}\right]^2-\left[x^{\frac{3}{2}}\right]^2}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]}}{\Delta x}\)
Simplify a bit:\(\lim _{x\rightarrow 0}\ \frac{\left[\left(x+\Delta x\right)^{\frac{3}{2}}\right]^2-\left[x^{\frac{3}{2}}\right]^2}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]\Delta x}\)
and some more...\(\lim _{x\rightarrow 0}\ \frac{\left(x+\Delta x\right)^3-x^3}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]\Delta x}\)
Now we have a
difference of cubes on this top
So we still can't take the limit. Here's what the difference of squares formula looks like:
Recall the difference of squares formula:\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
Once again, we'll set a and b:
\(a=\left(x+\Delta x\right)\)
\(b=x\)
Apply the new a and b into the formula:
\(a^3-b^3=\left(x+\Delta x-x\right)\left(\left(x+\Delta x\right)^2+\left(x+\Delta x\right)x+x^2\right)\)
Substitute the right side back into the limit!
\(\lim _{x\rightarrow 0}\ \frac{\left(x+\Delta x-x\right)\left(\left(x+\Delta x\right)^2+\left(x+\Delta x\right)x+x^2\right)}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]\Delta x}\)
Rearrange and simplify:\(\lim _{x\rightarrow 0}\ \frac{\left(\Delta x\right)\left(\left(x+\Delta x\right)^2+x^2+\Delta x\cdot x+x^2\right)}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]\Delta x}\)
The delta x now cancels out:
\(\lim _{x\rightarrow 0}\ \frac{\left(x+\Delta x\right)^2+x^2+\Delta x\cdot x+x^2}{\left[\left(x+\Delta x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}\right]}\)
Apply the limit as x approaches 0:
\(=\frac{x^2+x^2+x^2}{\left(x\right)^{\frac{3}{2}}+x^{\frac{3}{2}}}\)
Simplify more:
\(=\frac{3x^2}{2\left(x\right)^{\frac{3}{2}}}\)
Simplify more to eventually get\(\frac{3\sqrt{x}}{2}\)
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