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Differentiate with respect to x the functions in exercise x^{x}+x^{2}+a^{x}+a^{a}, for some fixed a>0 and x>0.
CBSE | Class 12 | Continuity and Differentiability | Maths | Miscellaneous | NCERT
Differentiate with respect to x the functions in exercise x^{x}+x^{2}+a^{x}+a^{a}, for some fixed a>0 and x>0.

Solution:

Let’s consider y=x^{x}+x^{a}+a^{x}+a^{a}

Now applying derivation:

\frac{d y}{d x}=\frac{d}{d x} x^{x}+\frac{d}{d x} x^{a}+\frac{d}{d x} a^{x}+\frac{d}{d x} a^{a}

=\frac{d}{d x} x^{x}+a x^{\alpha-1}+a^{x} \log a+0………….(i)

First term from equation (i):

\frac{d}{d x}\left(x^{x}\right), Let’s consider u=x^{x}

\log u=\log x^{x}=x \log x

\frac{d}{d x} \log u=\frac{d}{d x}(x \log x)

\frac{1}{u} \frac{d u}{d x}=x \frac{d}{d x} \log x+\log x \frac{d}{d x} x

\frac{1}{u} \frac{d u}{d x}=x \frac{1}{x}+\log x .1

=1+\log x

Which implies,

\frac{d u}{d x}=u(1+\log x)

Now, substituting the value of u back:

\frac{d}{d x} x^{x}=x^{x}(1+\log x)……………..(ii)

By using the eq. (ii) in (i), we get

\frac{d y}{d x}=x^{x}(1+\log x) a x^{a-1}+a^{x} \log a

i

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