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Calculus Examples
xsec(x)xsec(x)
Step 1
Step 1.1
Rewrite xsec(x)xsec(x) as eln(xsec(x))eln(xsec(x)).
ddx[eln(xsec(x))]ddx[eln(xsec(x))]
Step 1.2
Expand ln(xsec(x))ln(xsec(x)) by moving sec(x)sec(x) outside the logarithm.
ddx[esec(x)ln(x)]ddx[esec(x)ln(x)]
ddx[esec(x)ln(x)]ddx[esec(x)ln(x)]
Step 2
Step 2.1
To apply the Chain Rule, set u as sec(x)ln(x).
ddu[eu]ddx[sec(x)ln(x)]
Step 2.2
Differentiate using the Exponential Rule which states that ddu[au] is auln(a) where a=e.
euddx[sec(x)ln(x)]
Step 2.3
Replace all occurrences of u with sec(x)ln(x).
esec(x)ln(x)ddx[sec(x)ln(x)]
esec(x)ln(x)ddx[sec(x)ln(x)]
Step 3
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=sec(x) and g(x)=ln(x).
esec(x)ln(x)(sec(x)ddx[ln(x)]+ln(x)ddx[sec(x)])
Step 4
The derivative of ln(x) with respect to x is 1x.
esec(x)ln(x)(sec(x)1x+ln(x)ddx[sec(x)])
Step 5
Combine sec(x) and 1x.
esec(x)ln(x)(sec(x)x+ln(x)ddx[sec(x)])
Step 6
The derivative of sec(x) with respect to x is sec(x)tan(x).
esec(x)ln(x)(sec(x)x+ln(x)(sec(x)tan(x)))
Step 7
Step 7.1
Apply the distributive property.
esec(x)ln(x)sec(x)x+esec(x)ln(x)(ln(x)(sec(x)tan(x)))
Step 7.2
Combine esec(x)ln(x) and sec(x)x.
esec(x)ln(x)sec(x)x+esec(x)ln(x)ln(x)sec(x)tan(x)
Step 7.3
Reorder terms.
esec(x)ln(x)sec(x)tan(x)ln(x)+esec(x)ln(x)sec(x)x
esec(x)ln(x)sec(x)tan(x)ln(x)+esec(x)ln(x)sec(x)x