Derivative help - calculus!?

log(3)*3^x+log(2)*2^x

ln(2)*(2^x)+ln(3)*(3^x) is the derivative, just plug in -.1

if the value of x is predetermined

than all you can do is substitute

the known value into the equation

2^(-0.1)+3^(-0.1)

Raise 2 to the -0.1th power.

0.933+3^(-0.1)

Raise 3 to the -0.1th power.

0.933+0.896

Add 0.896 to 0.933 to get 1.829.

1.829

However

they want you to determine the

derivative first then plug in the

x=-0.1

2^(x)+3^(x)

The derivative of 2^(x) is 2^(x)*ln(2).

(d)/(dx) 2^(x)+3^(x)=

2^(x)*ln(2)+(d)/(dx) 3^(x)

The derivative of 3^(x) is 3^(x)*ln(3).

(d)/(dx) 2^(x)+3^(x)=

2^(x)*ln(2)+3^(x)*ln(3)

Multiply 2^(x) by ln(2) to get 2^(x)ln(2).

(d)/(dx) 2^(x)+3^(x)=

2^(x)ln(2)+3^(x)*ln(3)

Multiply 3^(x) by ln(3) to get 3^(x)ln(3).

(d)/(dx) 2^(x)+3^(x)=

2^(x)ln(2)+3^(x)ln(3)

The derivative of 2^(x)+3^(x) is

2^(x)ln(2)+3^(x)ln(3).

now plug in the -0.1 into x

f(x) = 2^x + 3^x

f ' (x) = 2^x * ln(2) + 3^x * ln(3)

so f ' (-0.1) = 2^(-0.1) ln(2) + 3^(-0.1) * ln(3)

= 1.631