Question

Differentiate :
\(y = \{ f(x) \} ^ {\phi (x) } \)

Answer 1

@Miracrown and @vishweshshrimali5

Answer 2

{..}= fractional part?

Answer 3

aha logarithmic differentiation

Answer 4

Yep. Logarithmic Differentiation.

Answer 5

Hint use log both sides and then differentiate both sides (use product rule on RHS)

Answer 6

Okay yes... that's one method. But, there are 2 more methods.

Answer 7

Any ideas for those methods?

and no @cody_123 - Not fractional part.

Answer 8

\[\phi(x)*\ln y=\ln f(x)\]

Answer 9

Well I can surely think of one method as well;

I am going to first differentiate by assuming f(x) is constant and power is a function then I am going to assume that phi(x) is constant f(x) is some variable function

Answer 10

now use product rule of Lhs and chain rule of RHS

Answer 11

@cody_123
the equation is wrong

Answer 12

\[\large y = \{ f(x) \} ^ {\phi (x) } = e^{ \phi (x) \ln f(x)} \]

Answer 13

Yes @ganeshie8 's method is also an other method.

Answer 14

oh yes sorry \[\ln y=\phi(x)\ln f(x)\]

Answer 15

yeah

Answer 16

Okay, so, third method is Partial fractions?

Answer 17

nope

Answer 18

From Mathematica 9 Home Edition.
\[\frac{\partial f(x)^{\phi (x)}}{\partial x}=f(x)^{\phi (x)} \left(\frac{\phi (x) f'(x)}{f(x)}+\log (f(x)) \phi '(x)\right) \]

Answer 19

That's a formula worth learning for objective questions.

Answer 20

differentiate RHS WITH CHAIN RULE

Answer 21

and product rule

Answer 22

Okay, well, i was just confirming the methods. I have a question based on this. Will post soon.

Thanks

Answer 23

Actually you can understand this formula very easily if you pay attention on my second method.

1) If \(y = a^{f(x)}\), then,

\(y' = a^{f(x)}*ln(a)*f'(x)\)

2) If \(y = f(x)^a\), then,

\(y' = a*f(x)^{a-1}*f'(x) = a*f(x)^{a}*f'(x)*\cfrac{1}{f(x)}\)

So, using these two formulae above,

for \(y = {f(x)}^{g(x)}\),

\(y' = {f(x)}^{g(x)}*ln(f(x))*g'(x) + {f(x)}^{g(x)}*g(x)*f'(x)*\cfrac{1}{f(x)}\)

\(y' = {f(x)}^{g(x)}[ln[f(x)]*g'(x) + \cfrac{f'(x)*g(x)}{f(x)}]\)

Which is the formula @robtobey mentioned.

Answer 24

a very nice application of partial derivatives :)

Answer 25

Yeah thanks @ganeshie8

Answer 26

Thanks a lot @vishweshshrimali5 .