How do you find the derivative of #f(x) = 4/(sqrt(x))# using the limit definition?

Question

Emily Ammerman · Accepted Answer

#dy/dx=-2x^(-3/2)#

Let #y=4/sqrtx#

Replace #y# with #(y+Deltay)# and #x# with #(x+Deltax)#

#y=4/sqrtx#

#y+Deltay=4/(sqrt(x+Deltax))#

Subtract #y# and its equivalent #4/sqrtx# from both sides of the equation

#y+Deltay-y=4/(sqrt(x+Deltax))-4/sqrtx#

#Deltay=4/(sqrt(x+Deltax))-4/sqrtx#

Combine the fractions using the LCD#=sqrt(x)*sqrt(x+Delta x)#

#Deltay=4/(sqrt(x+Deltax))-4/sqrtx#

#Deltay=(4sqrtx-4sqrt(x+Deltax))/(sqrtxsqrt(x+Deltax))#

Factor out the #4#

#Deltay=(4(sqrtx-sqrt(x+Deltax)))/(sqrtxsqrt(x+Deltax))#

Multiply the numerator and denominator by #(sqrtx+sqrt(x+Deltax))#

#Deltay=(4(sqrtx-sqrt(x+Deltax)))/(sqrtxsqrt(x+Deltax))*((sqrtx+sqrt(x+Deltax)))/((sqrtx+sqrt(x+Deltax)))#

#Deltay=(4((sqrtx)^2-(sqrt(x+Deltax))^2))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

#Deltay=(4(x-(x+Deltax)))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

#Deltay=(4(x-x-Deltax))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

#Deltay=(4(-Deltax))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

Divide both sides by #Deltax#

#(Deltay)/(Deltax)=(4(-Deltax))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))*1/(Deltax)#

#(Deltay)/(Deltax)=(4(-cancel(Deltax)))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))*1/cancel(Deltax)#

#(Deltay)/(Deltax)=(4(-1))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

Take the limit of both sides as #Deltax rarr 0#

#dy/dx=lim_(Deltax rarr 0)(Deltay)/(Deltax)=lim_(Deltax rarr 0)(4(-1))/((sqrtxsqrt(x+Deltax))*(sqrtx+sqrt(x+Deltax)))#

#dy/dx=(4(-1))/((sqrtxsqrt(x+0))*(sqrtx+sqrt(x+0)))#

#dy/dx=(-4)/(sqrtxsqrtx*(sqrtx+sqrtx))#

#dy/dx=(-4)/(x*(2sqrtx))#

#dy/dx=(-2)/x^(3/2)#

#dy/dx=-2x^(-3/2)#

God bless....I hope the explanation is useful.

Ezra Adams · Answer

undefined To find the derivative of $ f(x) = \frac{4}{\sqrt{x}} $ using the limit definition:

1. Start with the definition of the derivative:
$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

2. Substitute $ f(x) = \frac{4}{\sqrt{x}} $ into the formula:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{4}{\sqrt{x+h}} - \frac{4}{\sqrt{x}}}{h} $$

3. Rationalize the numerator by multiplying both the numerator and the denominator by the conjugate of the numerator:
$$ f'(x) = \lim_{h \to 0} \frac{\frac{4(\sqrt{x} - \sqrt{x+h})}{(\sqrt{x} \cdot \sqrt{x+h})}}{h} $$

4. Simplify the expression:
$$ f'(x) = \lim_{h \to 0} \frac{4(\sqrt{x} - \sqrt{x+h})}{h \cdot (\sqrt{x} \cdot \sqrt{x+h})} $$

5. Cancel out $ h $ in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} \frac{4(\sqrt{x} - \sqrt{x+h})}{h \cdot (\sqrt{x} \cdot \sqrt{x+h})} \times \frac{\frac{1}{h}}{\frac{1}{h}} $$
$$ f'(x) = \lim_{h \to 0} \frac{4(\sqrt{x} - \sqrt{x+h})}{(\sqrt{x} \cdot \sqrt{x+h})} $$

6. Apply the limit:
$$ f'(x) = \frac{4(\sqrt{x} - \sqrt{x})}{(\sqrt{x} \cdot \sqrt{x})} $$
$$ f'(x) = \frac{4 \cdot 0}{x} $$
$$ f'(x) = 0 $$

Therefore, the derivative of $ f(x) = \frac{4}{\sqrt{x}} $ is $ f'(x) = 0 $.