# Find the exact value of the gradient of the curve y=e^(2-x)ln(3x-2) at the point on the curve where x=2.

This is a typical gradient question for A2 papers, as it requires use of the Product Rule and Chain Rule, as well as knowledge of e^x.

Firstly, we can see that we will need to use the Product Rule as e^2-x and ln(3x-2) are being multiplied together. As the Product Rule is dy/dx= u.dv/dx + v.du/dx, we will also need to find out the differentials of e^2-x and ln(3x-2), where we let e^2-x = u and ln(3x-2) = v.

__ u=e^2-x __ This is 2 a combination of operations, meaning we will need the Chain Rule to obtain du/dx.If we let 2-x=t, then u=e^t. Differentiating 2-x mans that dt/dx= -1 and differentiating e^t means that du/dt remains e^t. The Chain Rule is du/dx=dt/dx*du/dt, meaning that du/dx=-1*e^t, and substituting t back into the eqution leaves us with

**du/dx=-e^2-x**.

__ v=ln(3x-2)__ Again, we have 2 operations meaning the Chain Rule will be used to obtain dv/dx. If we let 3x-2=m, then v=ln(m). Differentiating 3x-2 means dm/dx=3, and differentiating ln(m) means dv/dm=1/m. Again, we will need to use the Chain Rule, dv/dx=dm/dx*dv/dm, meaning dv/dx=3*1/m. Substituting our m back into this equation leaves us with

**dv/dx=**

**3/3x-2.**

As we have now obtained our du/dx and our dv/dx, we cn use the Product Rule to find the gradient of our original function. The Product Rule is dy/dx=u.dv/dx+v.du/dx, meaning dy/dx=(e^2-x)(3/3x-2)+(ln(3x-2))(-e^2-x).

As we have an x value given to us in the question, we do not need to simplify this eqution and can cimply substitute our x value of 2 into this equation to find the gradient of the curve at that point. With our substituted x value of 2, dy/dx=(e^0)(3/4)+(ln(4))(-e^0)=(1)(3/4)+(ln(4))(-1). Therefore, our final answer is **dy/dx=3/4-ln(4)** at on the point on the curve where x=2. Using our Laws of Logarithms, this can also be written as **dy/dx=3/4+ln(4)^-1=3/4+ln(1/4)**, and either answer would be ccepted in the exam.