Derivative of ln(x+1)\ln\left(x + 1\right)

The calculator will find the derivative of ln(x+1)\ln\left(x + 1\right), with steps shown.

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Your Input

Find ddx(ln(x+1))\frac{d}{dx} \left(\ln\left(x + 1\right)\right).

Solution

The function ln(x+1)\ln\left(x + 1\right) is the composition f(g(x))f{\left(g{\left(x \right)} \right)} of two functions f(u)=ln(u)f{\left(u \right)} = \ln\left(u\right) and g(x)=x+1g{\left(x \right)} = x + 1.

Apply the chain rule ddx(f(g(x)))=ddu(f(u))ddx(g(x))\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right):

(ddx(ln(x+1)))=(ddu(ln(u))ddx(x+1)){\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)}

The derivative of the natural logarithm is ddu(ln(u))=1u\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}:

(ddu(ln(u)))ddx(x+1)=(1u)ddx(x+1){\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right)

Return to the old variable:

ddx(x+1)(u)=ddx(x+1)(x+1)\frac{\frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}

The derivative of a sum/difference is the sum/difference of derivatives:

(ddx(x+1))x+1=(ddx(x)+ddx(1))x+1\frac{{\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}

The derivative of a constant is 00:

(ddx(1))+ddx(x)x+1=(0)+ddx(x)x+1\frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x\right)}{x + 1} = \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)}{x + 1}

Apply the power rule ddx(xn)=nxn1\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1} with n=1n = 1, in other words, ddx(x)=1\frac{d}{dx} \left(x\right) = 1:

(ddx(x))x+1=(1)x+1\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x + 1} = \frac{{\color{red}\left(1\right)}}{x + 1}

Thus, ddx(ln(x+1))=1x+1\frac{d}{dx} \left(\ln\left(x + 1\right)\right) = \frac{1}{x + 1}.

Answer

ddx(ln(x+1))=1x+1\frac{d}{dx} \left(\ln\left(x + 1\right)\right) = \frac{1}{x + 1}A