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Differentiate, with respect to x, (a) $e^{3x} + ext{ln} 2x,$ (b) $(5 + x^2)^{ rac{1}{2}}.$ - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 4

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Question 4

Differentiate,-with-respect-to-x,--(a)-$e^{3x}-+--ext{ln}-2x,$--(b)-$(5-+-x^2)^{-rac{1}{2}}.$-Edexcel-A-Level Maths Pure-Question 4-2006-Paper 4.png

Differentiate, with respect to x, (a) $e^{3x} + ext{ln} 2x,$ (b) $(5 + x^2)^{ rac{1}{2}}.$

Worked Solution & Example Answer:Differentiate, with respect to x, (a) $e^{3x} + ext{ln} 2x,$ (b) $(5 + x^2)^{ rac{1}{2}}.$ - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 4

Step 1

(a) $e^{3x} + ext{ln} 2x$

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Answer

To differentiate the expression, we use the rules of differentiation for exponential and logarithmic functions.

First, consider the term e3xe^{3x}. The derivative of ekxe^{kx} with respect to xx is kekxke^{kx}. Thus, we have:

ddxe3x=3e3x.\frac{d}{dx} e^{3x} = 3e^{3x}.

Now, for the term extln2x ext{ln} 2x, we need to apply the chain rule. The derivative of extlnu ext{ln} u is 1ududx\frac{1}{u} \cdot \frac{du}{dx}, where u=2xu = 2x:

ddxln2x=12x2=1x.\frac{d}{dx} \text{ln} 2x = \frac{1}{2x} \cdot 2 = \frac{1}{x}.

Combining both results, the final answer for part (a) is:

dydx=3e3x+1x.\frac{dy}{dx} = 3e^{3x} + \frac{1}{x}.

Step 2

(b) $(5 + x^2)^{\frac{1}{2}}$

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Answer

For part (b), we will use the chain rule to differentiate the expression. Let:

y=(5+x2)12.y = (5 + x^2)^{\frac{1}{2}}.

Using the chain rule, we differentiate as follows:

  1. The outer function is u12u^{\frac{1}{2}}, where u=5+x2u = 5 + x^2. The derivative of u12u^{\frac{1}{2}} is 12u12\frac{1}{2} u^{-\frac{1}{2}}.
  2. The derivative of uu with respect to xx is ddx(5+x2)=2x\frac{d}{dx}(5 + x^2) = 2x.

Putting it all together, we have:

dydx=12(5+x2)122x=x(5+x2)12.\frac{dy}{dx} = \frac{1}{2}(5 + x^2)^{-\frac{1}{2}} \cdot 2x = \frac{x}{(5 + x^2)^{\frac{1}{2}}}.

Thus, the final answer for part (b) is:

dydx=x(5+x2)12.\frac{dy}{dx} = \frac{x}{(5 + x^2)^{\frac{1}{2}}}.

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