NIMCET 2018 Question Paper with solutions

© examsiri.com

Question : 20 of 120

Marks: +1, -0

Differential coefficient of

log_{10}

w. r. t

\log_x

10 is

- $(logx)^2/(log10)^2$
$(log_{10}x)^2/(log 10)^2$
$(log_{x} 10)^2/(log 10)^2$
- $(log10)^2/(logx)^2$

Solution:

Let
u =

\log_{10}

x and v =

\log_x

10
u =

{log_e x}/{log_e 10}

, v =

{log_e 10}/{log_e x}

{du}/{dx}

=

1/{log_e 10} . 1/x

[differentiating w.r.t x]
and

{dv}/{dx}

=

log_e

10 .

1/(log_e x)^2 . 1/x

Now ,

{du}/{dv}

=

{{du}/{dx}}/{{dv}/{dx}}

=

{1/{log_e 10} . 1/x}/{-{log_e 10}/(log_e x)^2 . 1/x}

= -

1/{log_e 10} . 1/x . x/{log_e 10} (log_e x)^2

= -

(logx)2/(log 10)^2

© examsiri.com

Go to Question:

Prev Question

Next Question