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Question: How do you differentiate \(\ln {{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}\)?...

How do you differentiate ln⁡(x2+1)12\ln {{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}?

Explanation

Solution

Now we are given with a composite function. Hence we will use chain rule of differentiation to solve the differential. Now using chain rule of differentiation we have differentiation of the function f(g(h(x)))f\left( g\left( h\left( x \right) \right) \right) as f′(g(h(x)))g′(h(x)).h′(x)f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right). Now we will find the differentiation of each function and hence substitute the values to find the differentiation of the given equation. To do so we have d(ln⁡x)dx=1x\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}, d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} and d(x)dx=12x\dfrac{d\left( \sqrt{x} \right)}{dx}=\dfrac{1}{2\sqrt{x}}

Complete step-by-step solution:
Now consider the given function ln⁡(x2+1)12\ln {{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}.
The given function is a composite function of the form f(g(h(x)))f\left( g\left( h\left( x \right) \right) \right) .
Now we know that to differentiate a composite function we use chain rule of differentiation.
Now using chain rule of differentiation we have d(f(g(x)))dx=f′(g(x)).g′(x)\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right) where f′(x)=d(f(x))dxf'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx} .
Now in the above function we have f(x)=ln⁡xf\left( x \right)=\ln x, g(x)==x12g\left( x \right)=={{x}^{\dfrac{1}{2}}} and h(x)=x2+1h\left( x \right)={{x}^{2}}+1.
Now consider f(x)=ln⁡xf\left( x \right)=\ln x
Now we know that the differentiation of ln⁡x\ln x is given by 1x\dfrac{1}{x} .
Hence we have f′(x)=1xf'\left( x \right)=\dfrac{1}{x} .
Now consider g(x)=x12g\left( x \right)={{x}^{\dfrac{1}{2}}}
Now differentiating we get, g′(x)=12x12g'\left( x \right)=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}}
And since h(x)=x2+1h\left( x \right)={{x}^{2}}+1 we have h′(x)=2xh'\left( x \right)=2x .
Now according to chain rule we have differentiation of the function f(g(h(x)))f\left( g\left( h\left( x \right) \right) \right) as
⇒f′(g(h(x)))g′(h(x)).h′(x)\Rightarrow f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right)
Hence on substituting the values in the given formula we have,
⇒1(x2+1)12.12(x2+1)12.(2x)\Rightarrow \dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}}.\dfrac{1}{2{{\left( {{x}^{2}}+1 \right)}^{\dfrac{1}{2}}}}.\left( 2x \right)
Now we know that xmxn=xm+n{{x}^{m}}{{x}^{n}}={{x}^{m+n}} . Hence using this and simplifying we get,
⇒xx2+1\Rightarrow \dfrac{x}{{{x}^{2}}+1}
Hence the differentiation of the given function is given by xx2+1\dfrac{x}{{{x}^{2}}+1}.

Note: Now note that we can use chain rule for a complete series of composition of function. The formula for chain rule is d(f(g(x)))dx=f′(g(x)).g′(x)\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right). Hence when we have a series of compositions we will use this formula inductively to find the formula for n number of compositions. Hence we get differentiation of the function f(g(h(x)))f\left( g\left( h\left( x \right) \right) \right) as
f′(g(h(x)))g′(h(x)).h′(x)f'\left( g\left( h\left( x \right) \right) \right)g'\left( h\left( x \right) \right).h'\left( x \right) . Also note that composition is not multiplication of the functions.