Derivatives of logarithms and exponentials

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General formula:

$d/dx(lnx)=1/x$

$d/dx(e^{x})=e^{x}$

Examples:

Q: Find $f'(x)$ of $f(x)=x^{3}+lnx^{5}$

A:

f'(x)=d/dx(x³+ln x⁵)

=d/dx(x³)+d/dx(ln x⁵) =3x²+5/x

Note: d/dx(ln x⁵)=5/x

General Formula:

d/dx(e^cx)= ce^cx

Example:

Q1: find f'(x),if

(1) f(x)=5e^-2x

(2) f(x)=e^-x/5

A1:

(1) f'(x)=d/dx (5e^-2x)

=5d/dx (e^-2x)

=5*(-2)e^-2x

=-10e^-2x

(2) f'(x)=d/dx(e-5/x)

=-1/5 e^-5/x

Q2: find f'(x) if f(x)=3^x

A2:

f'(x)=d/dx(3^x)

=d/dx(e^ln3x)

= d/dx(e^xln3)

=ln3 * e^xln3

=ln3* 3^x

General formula:

d/dx(bx)= (lnb)*bx

**Example:**

Q1:if at some time you have $2000 in an account paying an interest rate of 6% (continuous compounding), what is the instantaneous rate of change in your money account?

Answer

Recall: A= Pe^rt

dA/dt= d/dt(P^rt)= P d/dt(e^rt)=Pre^rt

=r A(t)

=2000*6%=120

Q2: Find the point where the graph of y=3e^5x+2 has a slope of 5

Answer: Slope= dy/dx=d/dx(3e^5x+2)

= 3d/dx(e^5x)+0

=15e^5x+0

=15 e^5x

To find the point we set

dy/dx =5= 15e^5x

e^5x=1/3

ln(e^5x)=ln(1/3)

5x= -ln3

x=(-1/5)*ln3

The y coordinate is y=3e^ln(1/3)/5+2

=3e^ln(1/3)+2

=3*1/3+2=3

So the point is ((-1/5)ln3,3)

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