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Question: How do you find the derivative of \[f(x) = \sqrt {1 + 3{x^2}} \] ?...

How do you find the derivative of f(x)=1+3x2f(x) = \sqrt {1 + 3{x^2}} ?

Explanation

Solution

We simply use chain rule of differentiation to calculate the derivative of the given function with respect to the variable ‘x’. First differentiate the power of the function, then differentiate the logarithm part of the function and in the end differentiate the innermost bracket.

  • General formula of differentiation isddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}
  • Chain rule of differentiation:ddxg(f(x))=ddxg(f(x))×ddxf(x)\dfrac{d}{{dx}}g\left( {f(x)} \right) = \dfrac{d}{{dx}}g(f(x)) \times \dfrac{d}{{dx}}f(x)

Complete step by step solution:
We are given the function f(x)=1+3x2f(x) = \sqrt {1 + 3{x^2}}
We have to find the derivative of the function with respect to ‘x’. So, we differentiate the function with respect to ‘x’. We use concepts of differentiation to calculate f(x)=1+3x2f(x) = \sqrt {1 + 3{x^2}} .
If we look at the function f(x)=1+3x2f(x) = \sqrt {1 + 3{x^2}} we can write
f(x)=g(x)f(x) = \sqrt {g(x)} and g(x)=1+3x2g(x) = 1 + 3{x^2}
So, using chain rule of differentiation we can solve
ddx(1+3x2)=df(x)dx\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = \dfrac{{df(x)}}{{dx}}
Now break the function on right hand side of the equation
ddx(1+3x2)=dfdg×dgdx\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dx}}
Now differentiate the functions on the right hand side of the equation accordingly.
ddx(1+3x2)=12(1+3x2)121×6x\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = \dfrac{1}{2}{\left( {1 + 3{x^2}} \right)^{\dfrac{1}{2} - 1}} \times 6x
Multiply the possible products on right hand side of the equation
ddx(1+3x2)=3x(1+3x2)12\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = 3x{\left( {1 + 3{x^2}} \right)^{ - \dfrac{1}{2}}}
Use the negative term in the power to write the reciprocal
ddx(1+3x2)=3x[11+3x2]\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = 3x\left[ {\dfrac{1}{{\sqrt {1 + 3{x^2}} }}} \right]
Multiply the terms on the right hand side of the equation.
ddx(1+3x2)=3x1+3x2\Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 + 3{x^2}} } \right) = \dfrac{{3x}}{{\sqrt {1 + 3{x^2}} }}
\therefore The derivative of the function f(x)=1+3x2f(x) = \sqrt {1 + 3{x^2}} is 3x1+3x2\dfrac{{3x}}{{\sqrt {1 + 3{x^2}} }}

Note: Many students make the mistake of solving the derivative wrong as they only differentiate the term inside the square root. Keep in mind square root itself is a separate function and we have to find differentiation of all functions that have variable ‘x’ in them. Also, students can write the square root function as power half of the function inside the square root. And in the end similarly write the negative power as reciprocal in the derivative of the function.