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How do you differentiate #y=b^x?

Answer 1

Charles Anctil

#dy/dx=b^x *ln b#

From the given #y=b^x#

#ln y=ln b^x#

#ln y=x*ln b#

#d/dx(ln y)=d/dx(x*ln b)#

#(1/y)*y'=(x*0+ln b)#

#y'=y*ln b#

#y'=b^x*ln b#

God bless.....I hope the explanation is useful.

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Answer 2

Cameron Alexander

To differentiate ( y = b^x ), where ( b ) is a constant base and ( x ) is the variable exponent, we use logarithmic differentiation. The steps are as follows:

Take the natural logarithm of both sides: ( \ln(y) = \ln(b^x) ).
Apply the logarithmic property ( \ln(b^x) = x \ln(b) ).
Differentiate both sides with respect to ( x ).
Use the chain rule when differentiating ( \ln(y) ).
Solve for ( \frac{{dy}}{{dx}} ).

The result will be:

[ \frac{{dy}}{{dx}} = \frac{{d}}{{dx}}(b^x) = b^x \cdot \ln(b) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.