Differentiate with respect to x

**duckegg** · June 15th 2009, 02:34 PM

Please attempt to show working where ever possible...

See the attached file for questions....

**Soroban** · June 15th 2009, 02:58 PM

Hello, duckegg!

Differentiate: . $(a)\;\;y \;=\;\sqrt{x^3+5}$

We have: . $y \;=\;\left(x^3+5\right)^{\frac{1}{2}}$

Chain Rule: . $\frac{dy}{dx} \;=\;\tfrac{1}{2}(x^3+5)^{-\frac{1}{2}}\cdot 3x^2 \;=\;\frac{3x^2}{2\sqrt{x^3+5}}$

$(b)\;\;y \;=\;x^3\sin2x$

Product Rule: . $\frac{dy}{dx}\;=\;x^3\cdot\cos2x\cdot2 + 3x^2\cdot\sin2x \;=\;2x^3\cos2x + 3x^2\sin2x$

$(c)\;\;y \:=\:\frac{e^{2x}}{x+3}$

Quotient Rule: . $\frac{dy}{dx}\;=\;\frac{(x+3)\cdot e^{2x}\cdot 2 - e^{2x}\cdot 1}{(x+3)^2} \;=\;\frac{e^{2x}(2x+6-1)}{(x+3)^2} \;=\;\frac{e^{2x}(2x+5)}{(x+3)^2}$

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